1 GROUPS Algebra Study Guide

1 Groups

1.1 Isomorphism Theorems

Thm: \First Isomorphism Theorem" If ' : G ! H is a homomorphism of groups, then ker' E G and G= ker' �=im'.1

Cor: Let ' : G! H be a homomorphism of groups.(i) ' is injective if and only if ker' = feg.(ii) [G : ker'] = jim'j.2

Thm: \Second/Diamond Isomorphism Theorem" Let G be a group, let A and B be subgroups of G and assume A �NG(B). Then:(i) AB is a subgroup of G,(ii) B E AB,(iii) A \B E A,(iv) AB=B �= A=(A \B).3

Thm: \Third Isomorphism Theorem" Let G be a group and let H;K E G. Then,(i) K=H E G=H,(ii) (G=H)=(K=H) �= G=K.4

Thm: \Fourth/Lattice Isomorphism Theorem" Let G be a group and let N E G. THen there is a bijection from the setof subgroups A of G which contain N onto the set of subgroups A = A=N of G=N . In particular, every subgroup of Gis of the form A=N for some subgroup A of G containing N (namely, its preimage in G under the natural projectionhomomorphism from G to G=N). This bijection has the properties for all A;B � G with N � A;N � B:(i) A � B if and only if A � B,(ii) if A � B, then [B : A] = [B;A],(iii) hA;B i =

A;B

�,

(iv) A \B = A \B,(v) A E G if and only if A E G.5

Thm: \Fundamental Theorem of Finitely Generated Abelian Groups" Let G be a �nitely generated abelian group, thenG �= Zr � Zn1 � Zn2 � � � � � Zns for r; n1; : : : ; ns 2 Z with r � 0, ni � 2 and ni j ni+1. Also, this form is unique.6

Def: The integer r in the FTFGAG is called the free rank or Betti number of G. The numbers n1; : : : ; ns are theinvariant factors of G. This form is the invariant factor decomposition of G.

Thm: \Elementary Divisors Form" Let G be an abelian group with jGj = n > 1. Let the unique prime factorization ofn be n = p�1

1 � � � p�kk . Then,(i) G �= A1 � � � � �Ak where jAij = p�ii .(ii) Each Ai has Ai �= Zp�1

i

� � � � � Zp�ti

where �1 � � � � � �t and �1 + � � �+ �t = �i.

(iii) The decomposition in (i) and (ii) is unique.7

Def: The integers p�ji in the elementary divisor form above are the elementary divisors of G.8

Def: If G is a �nite abelian group of type (n1; : : : ; nt), the integer t is called the rank of G. If G is a group, the exponentof G is the smallest n > 0 such that xn = e for all x 2 G (if no such integer exists, then the exponent of G is 1).9

1.2 Subgroup Technology

Def: Let ' : G ! H be a homomorphism with kernel K. The quotient group or factor group, G=K is the groupwhose elements are the �bers of ' with operation de�ned by representatives of the �bers.10

Def: Let H � G. If xHx�1 = H for all x 2 G, then H is normal in G, written H E G.Prop: Let N E G. The set of left cosets of N partitions G, and u; v 2 G have that uN = vN if and only if v�1u 2 N .11

Def: Let S be a subset of G. Let N = NS = fx 2 GjxSx�1 = Sg be called the normalizer of S in G.12

Def: Let S be a subset of G. Let Z = ZS = fx 2 Gjxyx�1 = y;8y 2 Sg be called the centralizer of S in G.13

Prop: Let G be a �nite group of order n > 1. Let a be an element of G n feg. Then the period of a divides n. If theorder of G is a prime number p, then G is cyclic and the period of any generator is equal to p.14

Prop: Let G be a group and let x 2 G and a 2 Z n f0g.(i) If jxj =1, then jxaj =1.(ii) If jxj = n <1;, then jxaj = n

gcd(n;a) .

(iii) In particular, if jxj = n < l1, a j n, then jxaj = na .

15

Derrick Stolee

1 GROUPS Algebra Study Guide 1.2 Subgroup Technology

Prop: Let H = hx i.(i) Assume jxj =1, then H = hxa i if and only if a = �1.(ii) Assume jxj = n <1, then H = hxa i if and only if gcd(a; n) = 1.16

Prop: Let G be a cyclic group. Then every subgroup of G is cyclic. If f is a homomorphism of G, then the image of fis cyclic.17

Prop: \Properties of Generators in Cyclic Groups"(i) An in�nite cyclic group has exactly two generators (if a is a generator, then a�1 is the only other generator).(ii) Let G be a �nite cyclic group of order n, and let x be a generator. The set of generators of G consists of thosepowers xv of x such that v is relatively prime to n.(iii) Let G be a cyclic group, and let a; b be to generators. Then there exists an automorphism of G mapping a ontob. Conversely, any automorphism of G maps a on some generator of G.(iv) Let G be a cyclic group of order n. Let d be a positive integer dividing n. Then there exists a unique subgroupof G of order d.(v) \Chinese Remainder Thm" Let G1; G2 be cyclic or orders m;n. If m;n are relatively prime, then G1�G2

�= Cmn.(vi) Let G be a �nite abelian group. If G is not cyclic, then there exists a prime p and a subgroup of G isomorphicto C � C where C is cyclic of order p.18

Thm: \Properties of Isomorphisms" Let � be an isomorphism from G to H. Then, for a; b 2 G; k; n 2 Z:(i) �(eG) = eH (ii) �(an) = �(a)n (iii) G = h a i , H = h�(a) i(iv) ab = ba, �(a)�(b) = �(b)�(a). (v) jaj = j�(a)j(vi) The equation xk = b has the same # of solutions in G as xk = �(b) in H.(vii) G is abelian/cyclic i� H is abelian/cyclic. (viii) ��1 is an isomorphism from H to G.(ix) If K � G, then �(K) � H.19

Cor:G � H if any of these conditions fail.Thm: Some automorphism groups:

(i) For every positive integer n, Aut(Zn) �= Z�n . jZ�n j = '(n) := the Euler ' function.

(ii) For p an odd prime, Aut(Zp) �= Zp�1.(iii) For p an odd prime and m = pn, Aut(Zm) is cyclic of order pn�1(p� 1).(iv) For m = 2n, Aut(Zm) �= Z2 � Zr where r = 2n�2 (not cyclic, but has a cyclic subgroup of index 2).(v) If p is prime and V is an abelian group of order pn such that pv = 0, for all v 2 V , then V is an n-dim vectorspace over the �eld Zp, and Aut(V ) �= GL(V ) �= GLn(Zp).(vi) If n 6= 6, then Aut(Sn) �= Inn(Sn) �= Sn (every automorphism is inner). (vii) [Aut(S6) : Inn(S6)] = 2 and thereexists an automorphism S6 ! S6.

20 (viii) Aut(D8) �= D8 and Aut(Q8) �= S4.21

(ix) If G = Cp � Cp for some p prime, G acts as a 2-dim v.s. over Fp. � 2 Aut(G) can take (a; 0) to any non-zeroelement (for a 6= 0) since every nonzero element has order p. Now, �(a; 0) is determined, so �(0; b) has p2� p choicesleft. jAut(G)j = (p2 � 1)(p2 � p).

Thm: If � is a homomorphism from G to H, a; b; g 2 G, n 2 Z, then(i) �(eG) = eH (ii) �(gn) = �(g)n (iii) If jgj <1, then j�(g)j j jgj. (iv) ker� E G.(v) �(a) = �(b), b�1a 2 ker�, a ker� = b ker�. (vi) If �(a) = b, then ��1(b) = fx 2 Gj�(x) = b g = a ker�.22

Thm: \Properties of Subgroups under Homomorphisms" If H � G and � : G! L a group homomorphism,(i) �(H) is a subgroup of L. (ii) If H is cyclic/abelian/normal, then �(H) is cyclic/abelian/normal.(iii) If j ker�j = n, then � is n-to-1 mapping from G to �(G). (iv) If ker� = feg, then � is injective.(v) If jHj = n, then j�(H)j divides n. (vi) If K � L, then ��1(K) = fx 2 Gj�(x) 2 K g is a subgroup of G.(vii) If K E L, then ��1(K) E G. (viii) If � surjective and ker� = feg, then � is isomorphism.23

Prop: \Recognition Theorem" Let H and K be subgroups of a group G.(i) If H \K = f1g, the product map p : H �K ! G de�ned by p(h; k) = hk is injective. im p = HK � G.(ii) If either H or K is a normal subgroup of G then the product sets HK and KH are equal, and HK � G.(iii) If H and K are both normal, H \K = f1g, and HK = G, then G �= H �K.24

(iv) jHKj = jHjjKjjH\Kj .

25 (v) HK is a subgroup if and only if HK = KH.26 (vi) The number of distinct ways to write

each element of HK in the form hk for some h 2 H and some k 2 K is jH \Kj.27

Def: Let H be a subgroup of G. A subgroup K � G is called a complement for H in G if G = HK and H \K = feg.28

Def: For an action of G on S, a �xed point of G is an element s 2 S such that xs = s for all x 2 G. The subset of Sconsisting of all elements xs for x 2 G is the orbit of s under G, denoted Gs. The set of all x 2 G such that xs = sis called the stabilizer of s in G, denoted Gs.

29

Def:An action of G on S is faithful if the map � : G ! Perm(S) is injective. the action is transitive if there is only

Derrick Stolee

1 GROUPS Algebra Study Guide 1.3 Sylow/Solvability Problems

one orbit.30

Prop: Let S be a G-set, and let s be an element of S. Let H be the stabilizer of s, and let Os be the orbit of s. There

is a natural bijective map G=H'�! Os de�ned by aH 7! as. This map is comparible with the operations of G in

the sense that '(gC) = g'(C) for every coset C and every element g 2 G.31

Prop: Let S be a G-set, and let s 2 S. Let s0 be an element in the orbit of s, say s0 = as. Then A. the set of elementsg of G such that gs = s0 is the left coset aGs = f ah j h 2 H g. B. The stabilizer of s0 is a conjugate subgroup of thestabilizer of s: Gs = aGsa

�1 = f aha�1 j h 2 H g.32

Prop: \LOIS" If G is a group operating on a set S, and s 2 S, then jGsj = [G : Gs].33

Prop: The number of conjugate subgroups to H is equal to the index of the normalizer in H. 34

Thm: \Orbit Decomposition Formula" Let a group G act on a set S. Let I be an indexing set whereGsi\Gsj 6= ; ) i = jfor all i; j 2 I. Then, jSj =

Pi2I(G : Gsi).

35

Cor: \Class Equation" Let G be a group acting on itself by conjugation. Let X be a system of distinct representativesfor the conjugacy classes of G. Then, jGj =

Px2X(G : Gx).

36

Thm: \Conjugation Action Rules" Let a; b 2 G and suppose aba�1 = br. Then, 8t; s,(i) abta�1 = brt. (ii) asbta�s = br

st.Def: If N;H are groups and : H ! Aut(N) is a homomorphism (written (h) = h), then there exists a semidirect

product group G = f (n; h) j n 2 N;h 2 H g with multiplication (n1; h1) � (n2; h2) = (n1 h1(n2); h1h2). Then also,N1 = f (n; eH)jn 2 N g E G and H1 = f (eN ; h)jh 2 H g � G, and G = N1H1; N1 \H1 = feg.

Def:A subgroup H � G is characteristic, written H char G, if �(H) = H for all � 2 Aut(G). A group G is character-istically simple if G has no non-trivial proper subgroups that are characteristic.37

Thm: (i) If H char K and K char G, then H char G. (ii) If H char K and K E G, then H E G.(iii) If H E G, then (H)� E G for any automorphism � of G.(iv) If H � K are subgroups of G such that H char G and K=H char G=H, then K char G.38

Thm: If H is a normal subgroup of G whose order and index are relatively prime, then H char G. 39

Thm:A characteristically simple group is the direct product of isomorphic simple groups. 40

Def:An elementary abelian p-group for p prime is Zp � � � � � Zp. 41

Thm: If H is a minimal normal subgroup of G, then either H is an elementary abelian p-group for some prime p or His the direct product of isomorphic nonabelian simple groups. 42

Prop: If A is any nonempty collection of subgroups of G, then the intersection of all members of A is also a subgroupof G.43

1.3 Sylow/Solvability Problems

Thm: \Sylow Theorems" Let G = pkm with p prime, and gcd(p;m) = 1.(I) If H is a p-subgroup of G, H � P , a p-Sylow subgroup of G.(II) All p-Sylow subgroups are conjugate in G.(III) Let sp denote the number of p-Sylow subgroups in G. sp � 1 (mod p), and sp j m.44

Lma: Let P be a p-Syslow subgroup of G. If Q is any p-subgroup of G, then Q \NG(P ) = Q \ P .45

Lma: Let H be a p-group acting on a �nite set S. Then:(a) The number of �xed points of H is � jSj (mod p).(b) If H has exactly one �xed point, then jSj � 1 (mod p).(c) If p j jSj, then the number of �xed points of H is � 0 (mod p).46

Thm: Let G be a �nite p-group. Then G is solvable. If jGj > 1, then G has non-trivial center.47

Cor: Let G be a p-group which is not of order 1. Then there is a normal, cyclic tower feg = G0 � G1 � � � � � Gn = G.Lma: Let G be a �nite group and let p be the smallest prime dividing the order of G. Let H be a subgroup of index p.

Then H is normal.48

Prop: Let p; q be distinct primes and let G be a group of order pq. Then G is solvable.49

Thm: \Little Cayley" For H � G, there exists a homomorphism � : G! Perm(G=H) by the action of left-multiplcationon the cosets of H. Then, ker � E H.50

Thm: Let H � G and have G act on the left cosets of H by left-multiplication. Then,(i) G acts transitively on the left cosets of H.(ii) The stabilizer of 1H is H.(iii) The kernel of the action is K = \x2GxHx

�1, and K is the largest normal subgroup of G contained in H.51

Lma: If G is a �nite group, H � G, and jHj = pr, p prime, then [G : H] � [NG(H) : H] (mod p).52

Derrick Stolee

1 GROUPS Algebra Study Guide 1.4 Symmetric Group

Lma: In a �nite group G, if sp 6� 1 (mod p2), then there are distinct p-Sylow subgroups P and R of G such that P \Ris of index p in both P and R (hence is normal in each).53

Prop: If the number of p-Sylow subgroups in G is given by sp = k, there exists a homomorphism � : G! Sk.Prop: If H � G and N E G, then there exists a homomorphism � : H ! Aut(N). The kernel, ker � = ZN , the

centralizer of H. And, G=ZN is isomorphic to a subgroup of Aut(N).54

Cor: For any H � G, NG(H)=ZH is isomorphic to a subgroup of Aut(H). In particular, G=Z(G) is isomorphic to asubgroup of Aut(G).55

Prop: \Methods for �nding subgroups" Consider a group G.(i) Find p-Sylow subgroups. (ii) NG(H) for H � G.(iii) NK(H) for H;K � G. (iv) HN for H � G, N E G. (v) H \K for H;K � G.

Prop: For p; q prime, p 6= q, groups of the following orders are solvable (requires proof):(i) pq (ii) p2q (iii) p2q2

Lma: Let S;N;H;K � G, N E G;S E H. Then(i) SN = NS & NH = HN . (ii) H \N E H. (iii) S \N E H;S \K E H \K. (iv) NS E NH.56

Lma: \Butter y Lemma" Let U; V be subgroups of a group. Let u E U and v E V . Then:(i) u(U \ v) E u(U \ V ).(ii) (u \ V )v E (U \ V )v.(iii) u(U \ V )=u(U \ v) �= (U \ V )v=(u \ V )v.57

Prop: Let G be a �nite group. An abelian tower of G admits a cyclic re�nement. Let G be a �nite solvable group.Then G admis a cyclic tower whose last element is feg.58

Thm: Let G be a group and H E G. Then G is solvable if and only if H and G=H are solvable.59

Def:A commutator in G is an element of the form xyx�1y�1. Let GC be the subgroup of G generated by thecommutators, called the commutator subgroup of G.60

Def: The commutator of two subgroups A;B � G is noted as [A;B] = f aba�1b�1 j a 2 A; b 2 B g.61

Prop: Let G be a group and GC be the commutator of G.(i) GC E G, G=GC abelian. (ii) N E G;G=N abelian implies GC � N . (So, GC is smallest normal subgroupwith an abelian factor group).(iii) G is solvable if and only if 9n so that (� � � (GC)C) � � � )C| {z }

n commutators

= feg.

(iv) For H � G, H E G if and only if [H;G] � H.(v) If ' : G! A is any homomorphism of G into an abelian group A, then there is a map : G=GC ! A such that � � = '.62

Lma: Suppose G is solvable, and has a solvable series of length r. Then, GCr�1

= feg (where GCn

denotes the nthcommutator of G).

Prop: Let H be a group which operates on a set S, and let U be a subset of S. Then H stabilizes U if and only if U isa union of H-orbits.63

Thm: The �nite group G is solvable if and only if for every divisor n of jGj such that gcd�n; jGjn

�= 1, G has a subgroup

of order n.64

Thm: \The Unusables!" Let G be a �nite group.(i) [Burnside] If jGj = paqb for p; q primes, then G is solvable.(ii) [Philip Hall] If for every prime p dividing jGj we factor the order of G as jGj = pam where gcd(p;m) = 1, and Ghas a subgroup of order m, then G is solvable (i.e. if for all prime p, G has a subgroup whose index equals the orderof a Sylow p-subgroup, then G is solvable { such subgroups are called Sylow p-complements).(iii) [Feit-Thompson] If jGj is odd, then G is solvable.(iv) [Thompson] If for every pari of elements x; y 2 G; hx; y i is a solvable group, then G is solvable.65

1.4 Symmetric Group

Prop: \Conjugation in Sn" For � = (i1 i2 : : : ir) and r-cycle in Sn and � 2 Sn, ����1 = (�(i1)�(i2) : : : �(ir)).

66

Thm: Recall automorphisms of Sn:(vi) If n 6= 6, then Aut(Sn) �= Inn(Sn) �= Sn (every automorphism is inner). (vii) [Aut(S6) : Inn(S6)] = 2 and thereexists an automorphism S6 ! S6

67.Thm: If n = 5, then Sn is not solvable.68

Prop:An is generated by the 3-cycles. If n � 5, all 3-cycles are conjugate in An. If n � 5, An is simple.

Derrick Stolee

1 GROUPS Algebra Study Guide 1.5 Free/Dual Groups

Prop: The permutation � 2 Sn is odd if and only if the number of cycles of even length in its cycle decomposition isodd.69

Def: If � 2 Sn is the product of disjoint cycles of length n1; n2; : : : ; nr with ni � ni+1, then the integers n1; n2; : : : ; nrare the cycle type of �.70

Prop: Two elements of Sn are conjugate in Sn if and only if they have the same cycle type. The number of conjugacyclasses of Sn equals the number of partitions of n, or the nth Bell number Bn.

71

Prop: (i) If G has no subgroup of index 2 and G � Sk, then G � Ak.(ii) If P is a p-Sylow subgroup of Sk for some odd prime p, then P is a p-Sylow subgroup of Ak and jNAk

(P )j =12 jNSk(P )j.

72

1.5 Free/Dual Groups

Def: Let G be a �nite abelian group. Let G = Hom(G;Q=Z).Fact: If f : G1 ! G2, then 9f : G2 ! G1 with f(�) = � � f .

Fact: \G1 �G2�= G1 � G2, for G1; G2; �nite abelian groups.

Fact: If G1f�! G2 is surjective, then G2

f�! G1 is surjective.

Def: Let X be a set. We construct F (X) de�ned to be the free group on X. If X = ;, F (X) = feg. If X 6= ;, let X�1

be disjoint from X, jXj = jX�1j. Let f : X ! X�1 be a bijection, label f(x) =: x�1. Add an element e =2 X [X�1

to S = X [X�1 [ feg. F (X) is the group of reduced concatenations of words over S.Def:A word on X is a sequence (a1; a2; : : : ) with each ai 2 S, and existing a k with an = e for all n � k. The contstant

(e; e; : : : ) is denoted ", the empty word.Def:A word is reduced if (i) 8x 2 X, x and x�1 are not adjacent. (ii) an = 1) ak = 1 for all k � n. Denote reduced

words as x�11 x�22 � � �x�nn where xi 2 X; � = �1, or " for the empty word.

Thm: If X 6= ;, then F (X) is a group and F = hX i (F is generated by X � F (X)).Thm: \Universal Property of Free Groups" there exists an i : X ! F (X) injective and: if G is a group and f : X ! G

is a set map then there exists a unique f : F ! G such that f � i = f .Cor: Every group G is the homomorphic image of a free group: Take F to be free on the generators of G. Then,

G �= F=W where W is the set of relations satis�ed by generators of G.Def: Let X be a set and Y a set of reduced words on X. A group G is said to be the group de�ned by the generators X

and relations w = e (with w 2 Y ) provided G �= F=N , where F is the free group on X and N the normal subgroupof F generated by Y . One says that hXjY i is a presentation of G.

Thm: Let X be a set, Y a set of reduced words on X and G = hXjY i. If H is any group such that H = hX i and Hsatis�es all the relations w = e (with w 2 Y ) then there is a homomorphism G! H.

1.6 Cardinality

Thm: Let A be an in�nite set. Then card(A�A) = card(A).73

Cor: Let A be an in�nite set, and let � be the set of �nite subsets of A. Then card(�) = card(A).Thm: Let A be an in�nite set, and T = f0; 1g. Let M be the set of all maps of A into T . Then, card(A) � card(M).74

Cor: Let A be an in�nite set, and S the set of all subsets of A. Then card(A) � card(S).75

1.7 Category Theory

Def:A category C consists of a collection of objects and for two objects A;B 2 C, a set Hom(A;B) is called the set ofmorphisms from A to B. For three objects A;B;C 2 C, a law of composition: Hom(B;C)�Hom(A;B)! Hom(A;C),satisfying the axioms:(CAT 1) Two sets Hom(A;B) and Hom(A0; B0) are disjoint unless A = A0 and B = B0, in which case they are equal.(CAT 2) For each object A of C, there is a morphism idA 2 Hom(A;A) which acts as right and left identity on theelements of Hom(A;B) and Hom(B;A), respectively.(CAT 3) The law of composition is associative.76

Def: The product of fAi gi2I in the category C is a P 2 C such that there exist maps �i : P ! Ai so that if there existsa D 2 C with maps 'i : D ! Ai there exists a unique ' : D ! P so that 'i = �i � '.

77

Def: The coproduct of fAi gi2I in the category C is a P 2 C such that there exist maps �i : Ai ! P so that if thereexists a D 2 C with maps i : Ai ! D there exists a unique : P ! D so that i = �i � .

78

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NOTES Algebra Study Guide NOTES

Def: For some X;Y; Z 2 C, with morphisms f : X ! Z and g : Y ! Z, the �ber product (or pull-back) of f and g in Cis an object P 2 C with morphisms p1 : P ! X, p2 : P ! Y such that f � p1 = g � p2 with the addition that if Q 2 Cwith morphisms � : Q! X and : Q! Y with f � � = g � , there exists a unique morphism p : Q! P such thatf � p1 � p = g � p2 � p.

Def: For some X;Y; Z 2 C, with morphisms f : Z ! X and g : Z ! Y , the �ber coproduct (or push-out) of f and g inC is an object P 2 C with morphisms p1 : X ! P , p2 : Y ! P such that p1 � f = p2 � g with the addition that ifQ 2 C with morphisms � : X ! Q and : Y ! Q with � � f = � g, there exists a unique morphism p : P ! Qsuch that p � p1 � f = p � p2 � g.

Prop: Products/Coproducts/Pull-backs/Push-outs, if they exist, are unique.Def: Let C;D be categories. A covariant functor F : C ! D is a rule so that 8C 2 C, F (C) 2 D, 8f 2 Arr(C), say

f : A! B;A;B 2 C, F (f) : F (A)! F (B). F satis�es(F1) 8C 2 C, F (idC) = idF (C).

(F2) If Af�! B

g�! C in C, then F (g � f) = F (g) � F (f).

Def:A contravariant functor G satis�es simiilar properties, except f : A! B ) G(f) : G(B)! G(A).

Artin: Algebra by Michael Artin.D&F: Abstract Algebra by Dummit & Foote.H: Algebra by Thomas W. Hungerford.Lang: Algebra by Serge Lang.

Notes

1D&F. Theorem 16. p.97.2D&F. Corollary 17. p. 97.3D&F. Theorem 18. p.97.4D&F. Theorem 19. p. 98.5D&F. Theorem 20. p. 99.6D&F. Theorem 3. p.158.7D&F. Theorem 5. p. 161.8D&F. p. 161.9D&F. p. 165.10D&F. p. 76.11D&F. Proposition 4. p. 80.12Lang, p. 14; D&F, p. 50.13Lang, p.14; D&F, p. 49.14Lang, Chapter 1, Prop 4.1. p.24.15D&F, Proposition 5, p. 57.16D&F, Proposition 6. p. 57.17Lang, Chapter 1, Prop 4.2. p.24.18Lang, Chapter 1, Prop 4.3. pp. 24-25.19Gallian 6.2, 6.3 pp. 126-12720Described in Dummit&Foote p. 222, #1021Gallian 6.5, p. 131; D&F, Prop 17 p. 136.22Gallian, 10.123Gallian, Theorem 10.224Artin, Section 2.8.25D&F. Proposition 13. p. 93.26D&F. Proposition 14. p. 94.27D&F. Proposition 8. p. 171.28D&F. p.180.29Lang, p. 15.30Lang, p. 15.31Artin, Section 5.6.32Artin, Section 5.6.33Lang, Chapter 1. Prop 5.1. p.15.34Lang, Chapter 1. Prop 5.2. p.15.

35Lang, p.29.36Lang, p.29.37S. Wiegand, Sept 28,2007; Gorenstein 196838S. Wiegand Sept 28,2007, handouts, Theorem 1.2.39S. Wiegand, Sept 28,2007, handouts, Theorem 1.3.40S. Wiegand, Sept 28,2007, handouts, Theorem 1.4.41S. Wiegand, Sept 28,2007; Gorenstein 196842S. Wiegand Sept 28,2007, handouts, Theorem 1.5.43D&F. Proposition 8. p. 62.44Lang, Chapter 1, Theorem 6.4, pp. 34-35.45D&F. Lemma 19. p. 140.46Lang, Chapter 1, Lemma 6.3, p. 34.47Lang, Chapter 1, Theorem 6.5, p. 3548Lang, Chapter 1, Lemma 6.7. p. 3649Lang, Chapter 1, Proposition 6.8. p.3650S. Wiegand Sept 10, 2007 handout.51D&F. Theorem 3. p.119.52Hungerford, Lemma 5.5 p. 94.53D&F. Lemma 13. p. 207.54D&F. Proposition 13. p.133.55D&F. Corollary 15. p.134.56S. Wiegand, Sept 10, 2007, handout. Lma for Butter y Lma.57Lang, Chapter 1. Lemma 3.3. pp. 20-21.58Lang, Chapter 1. Proposition 3.1. p.1859R. Wiegand, \Galois Theory Review" Lemma 3.6,3.7.60Lang, p.2061D&F. p.169.62D&F. Proposition 7. p.169.63Artin, Section 6.2.64D&F. p. 105.65D&F. Theorem 11. p. 196.66D&F. Proposition 10. p. 125.67Described in Dummit&Foote p. 222, #1068Lang, Chapter 1, Theorem 5.4. p. 31.69D&F. Proposition 25. p.110.70D&F. p.126.71D&F. Proposition 11. p. 126.72D&F. Proposition 12. p. 204.73Lang, Appendix 2, Theorem 3.6. p. 888.74Lang Appendix 2. Theorem 3.10, p.890.75Lang Appx 2. Cor 3.11, p. 891. Hungerford, Thm 8.5. p. 17.76Lang, p. 57.77Lang, p. 58.78Lang, p. 59.

Derrick Stolee

2 RINGS Algebra Study Guide

2 Rings

2.1 De�nitions

Def:A ring is a nonempty set R together with binary operations (addition and multiplication) such that(i) (R;+) is an abelian group. (ii) Multiplication is associative. (iii) a(b+c) = ab+ac; 8a; b; c 2 R (distributivelaw)If in addition, (iv) ab = ba for all a; b 2 R then R is commutative. If R contains an element 1R such that(v) 1Ra = a = a1R for all a 2 R, then R is said to be a ring with identity.79

Thm: Let R be a ring, a; b 2 R, and 0 be the additive identity. Then(i) 0a = a0 = 0; (ii) (�a)b = a(�b) = �(ab); (iii) (�a)(�b) = ab;

(iv) (na)b = a(nb) = n(ab) for all n 2 Z; (v)(Pni=1 ai)

�Pmj=1 bj

�=Pni=1

Pmj=1 aibj for all ai; bj 2 R.

80

Def:A nonzero element a in a ring R is a left [right ] zero divisor if there is a nonzero b 2 R such that ab = 0 [ba = 0].A zero divisor is an element of R which is both a left and right zero divisor.81

Def:An element a in a ring R with identity is said to be left [right ] invertible if there exists a c 2 R [b 2 R] such thatca = 1r [ab = 1R]. The element c [b] is called a left [right ] inverse of a. An element a 2 R that is both left and rightinvertible is said to be invertible or to be a unit.82

Thm: \Binomial Theorem" Let R be a ring with identity, n a positive integer, and a; b; a1; a2; : : : ; as 2 R.(i) if ab = ba, then (a+ b)n =

Pnk=0

�nk

�akbn�k;

(ii) If aiaj = ajai for all i; j, then (a1 + a2 + � � �+ as)n =

Pn!

(i!1)���(is!)

ai11 ai22 � � � a

iss ;

where the sum is over all s-tuples (i1; i2; : : : ; is) such that i1 + i2 + � � �+ is = n.83

Def: Let R and S be rings. A function f : R! S is a homomorphism of rings provided that for all a; b 2 R, f(a+ b) =f(a) + f(b) and f(ab) = f(a)f(b).84 A monomorphism [epimorphism,isomorphism] of rings is a homomorphism ofrings which is injective [surjective, bijective]. A monomorphism of rings R ! S is also called an embedding of R inS. An isomorphism R ! R is an automorphism of R. The kernel of a homomorphism is the kernel as a map ofadditive groups.85

Def: Let R be a ring. If there is a least positive integer n such that na = 0 for all a 2 R, then R is said to havecharacteristic n, written char R = n. If no such n exists, R is said to have characteristic zero.86

Thm: Let R be a ring with identity 1R and characteristic n > 0.(i) If ' : Z! R is the map given by m 7! m1R, then ' is a homomorphism of rings with kernel hn i = fknjk 2 Zg.(ii) n is the least positive integer such that n1R = 0.(iii) If R has no zero divisors (in particular if R is an integral domain) then n is prime.87

Thm: Every ring R may be embedded in a ring S with identity. The ring S (which is not unique) may be chosen to beeither characteristic zero or of the same characteristic as R.88

Def:An element of a ring is nilpotent if an = 0 for some n.89

2.2 Ideals

Def: Let R be a ring and S a nonempty subset of R that is closed under the operations of addition and multiplicationin R. If S is itself a ring under these operations then S is called a subring of R. A subring I of a ring R is called aleft ideal provided r 2 R and x 2 I implies rx 2 I. I is a right ideal provided r 2 R and x 2 I implies xr 2 I. I isan ideal (or two-sided ideal for emphasis) if it is both a left and right ideal.90

Def: The center of R is the set C = fc 2 R j cr = rc8r 2 Rg. C is a subring of R but may not be an ideal.91

Def: Every ring R has at least two ideals, R itself and the trivial ideal : the set 0 = f0Rg. An ideal I is called proper ifI 6= 0 and I 6= R.92

Thm:A nonempty subset I of a ring R is a left [right] ideal if and only if for all a; b 2 I and r 2 R:(i) a� b 2 I; and (ii) ra 2 I.93

Cor: Let fAi j i 2 Ig be a family of [left] ideals in a ring R. Then \i2IAi is also a [left] ideal.94

Def: Let X be a subset of a ring R. Let fAi j i 2 Ig be the family of all [left] ideals in R which contain X. Then\i2IAi is called the [left] ideal generated by X. This ideal is denoted hX i. The elements of X are called generatorsof the ideal hX i. If X is �nite, then hX i is �nitely generated. An element generated by a single element is called aprincipal ideal. A principle ideal ring is a ring in which every ideal is principal.95

Thm: Let R be a ring and a 2 R and X � R.(i) The principal ideal h a i consists of all elements of the form ra+ as+ na+

Pmi=1 riasi for r; s; ri; si 2 R;m 2 N,

Derrick Stolee

2 RINGS Algebra Study Guide 2.2 Ideals

and n 2 Z;(ii) If R has an identity, then h a i = f

Pni=1 riasi j ri; si 2 R;n 2 Ng;

(iii) If a is in the center of R, then h a i = fPmi=1 ria+ na j ri 2 R;m 2 N; n 2 Zg;

(iv) Ra = fra j r 2 Rg is a left ideal in R. If R has an identity, then a 2 Ra. De�ne aR similarly as a right ideal.(v) If R has an identity and a is in the center of R, then Ra = h a i = aR.(vi) If R has an identity and X is in the center of R, then the ideal hX i consists of all �nite sums r1a1+ � � �+ rnan,where n 2 N; ri 2 R; ai 2 X.96

Thm: Let A;A1; : : : ; An; B and C be [left] ideals in a ring R.(i) A1 +A2 + � � �+An and A1A2 � � �An are [left] ideals;(ii) (A+B) + C = A+ (B + C); (iii) (AB)C = ABC = A(BC);(iv) B(A1 +A2 + � � �+An) = BA1 + � � �+BAn and (A1 + � � �+An)C = A1C + � � �+AnC.

97

Thm: Let R be a ring and I an ideal of R. Then the additive quotient group R=I is a ring with multiplication given by(a+ I)(b+ I) = ab+ I. If R is commutative or has an identity, then the same is true of R=I.98

Thm: If f : R ! S is a homomorphism of rings, then the kernel of f is an ideal in R. Conversely, if I is an ideal in R,then the map � : R ! R=I given by r 7! r + I is an epimorphism of rings with kernel I. The map � is called thecanonical epimorphism (or projection) onto R=I.99

Thm: If f : R ! S is a homomorphism of rings and I is an ideal of R which is contains in the kernel of f , thenthere is a unique homomorphism of rings f : R=I ! S such that f(a + I) = f(a) for all a 2 R. im f = im f andker f = (ker f)=I. f is an isomorphism if and only if f is an epimorphism and I = ker f .100

Cor: \First Isomorphism Theorem of Rings" If f : R! S is a homomorphism of rings, then f induces an isomorphismof rings R= ker f �= im f .101

Cor: If f : R ! S is a homomorphism of rings, I is an ideal in R and J is an ideal in S such that f(I) � J , then finduces a homomorphism of rings f : R=I ! S=J , given by a + I 7! f(a) + J , f is an isomorphism if and only ifim f + J = S and f�1(J) � I. In particular, if f is an epimorphism such that f(I) = J and ker f � I, then f is anisomorphism.102

Thm: Let I and J be ideals in a ring R.(i) \Second Isomorphism Theorem" there is an isomorphism of rings I=(I \ J) �= (I + J)=J ;(ii) \Third Isomorphism Theorem" if I � J , then J=I is an ideal of R=I and there is an isomorphism of rings(R=I)=(J=I) �= R=J .103

Thm: \Ideal Correspondence Theorem" If I is an ideal in a ring R, then there is a one-to-one correspondence betweenthe set of all ideals of R which contain I and the set of all ideals of R=I, given by J 7! J=I. Hence every ideal inR=I is of the form J=I, where J is an ideal of R which contains I.104

Def:An ideal P in a ring R is said to be prime if P 6= R and for any ideals A;B in R, AB � P implies A � P orB � P .105

Thm: If P is an ideal in a ring R such that P 6= R and for all a; b 2 R ab 2 P implies a 2 P or b 2 P , then P Is prime.Conversely, if P is prime and R is commutative, then the reverse holds.106

Def:An ideal [left ideal] M in a ring R is maximal [left maximal ] if M 6= Rand for every [left] ideal N such thatM � N � R then either N =M or N = R.107

Prop: Let R be a commutative ring with an ideal I.(i) I is prime if and only if R=I is an integral domain.(ii) I is maximal if and only if R=I is a �eld.108

Thm: In a nonzero ring R with identity maximal [left] ideals always exist. In fact, every [left] ideal in R (except R itself)is contained in a maximal [left] ideal.109

Thm: If R is a commutative ring such that R2 = R (in particular if R has an identity), then every maximal ideal M inR is prime.110

Ex:Given a ring R with R2 = R, it is not true that every prime ideal is maximal: Consider 0 � Z, prime but notmaximal.111

Thm: Let fRi j i 2 Ig be a nonempty family of rings andQi2I Ri the direct product of the additive abelian groups Ri;

(i)Qi2I Ri is a ring with multiplication de�ned by faigi2Ifbigi2I = faibigi2I , called the (external) direct product

of the family fRig; (ii) If Ri has an identity [is commutative] for every i 2 I, thenQi2I Ri has an identity [is

commutative].(iii) for each k 2 I the canonical projection �k :

Qi!I Ri ! Rk given by faig 7! ak, is an epimorphism of rings;

(iv) for each k 2 I the canonical injection {k : Rk !Qi2I Ri, given by ak 7! faig where ai = 0 for i 6= k, is a

monomorphism of rings.112

Derrick Stolee

2 RINGS Algebra Study Guide 2.3 Domains and Factorization

Thm: Let fRi j i 2 Ig be a nonempty family of rings, S a ring and f'i : S ! Ri j i 2 Ig a family of homomorphisms ofrings. Then there is a unique homomorphism of rings ' : S !

Qi2I Ri such that �i' = 'i for all i 2 I. The ringQ

i2I Ri is uniquely determined up to isomorphism by this property. In other words,Qi2I Ri is a product in the

category of rings.113

Thm: Let A1; : : : ; an be ideals in a ring R such that (i) A1 + � � �+An = R and (ii) For each k, Ak \ (A1 + � � �+Ak�1 +Ak+1 + � � � + An) = 0. Then there is a ring isomorphism R �= A1 � � � � � An. R is said to be the (internal) directproduct of the ideals Ai.

114

Def: Let A be an ideal in a ring R and a; b 2 R. The element a is said to be congruent to b modulo A (denoted a � b(mod A)) if a� b 2 A. Thus, a � b (mod A), a� b 2 A, a+A = b+A:115

Thm: \Chinese Remainder Theorem" Let A1; : : : ; An be ideals in a ring R such that R2+Ai = R for all i and Ai+Aj = Rfor all i 6= j. If b1; : : : ; bn 2 R, then there exists b 2 R such that b � bi (mod Ai) for all i. Furthermore, b is uniquelydetermined up to congruence modulo the ideal A1 \A2 \ � � � \An.

116

Cor: Let m1;m2; : : : ;mn be positive integers such that gcd(mi;mj) = 1 for i 6= j If b1; : : : ; bn are any integers,then the system of congruences x � bi (mod mi) has an integral solution that is uniquely determined modulom = m1m2 � � �mn.

117

Cor: IfA1; : : : ; An are ideals in a ringR, then there is a monomorphism of rings � : R=(A1\� � �\An)! R=A1�� � ��R=An.If R2 +A = R for all i and Ai +Aj = R for all i 6= j then � is an isomorphism.118

Def:An element e in a ring R is said to be idempotent if e2 = e. Idempotent elements e1; : : : ; en in a ring R are saidto be orthogonal if eiej = 0 for i 6= j. An element of the center of the ring R is said to be central.119

2.3 Domains and Factorization

Def:A commutative ring R with identity 1R 6= 0 and no zero divisors is called an integral domain. A ring D withidentity 1D 6= 0 in which every nonzero element is a unit is called a division ring. A �eld is a commutative divisionring.120

Ex: The zero ideal in any integral domain is prime since ab = 0 if and only if a = 0 or b = 0.121

Thm: In a commutative ring R with identity 1R, an ideal P is prime if and only if the quotient ring R=P is an integraldomain.122

Thm: Let M be an ideal in a ring R with identity 1R 6= 0.(i) If M is maximal and R is commutative, then the quotient ring R=M is a �eld.(ii) If the quotient ring R=M is a division ring, then M is maximal.123

Cor: The following conditions on a commutative ring R with identity 1R 6= 0 are equivalent:(i) R is a �eld; (ii) R has no proper ideals; (iii) 0 is a maximal ideal in R;(iv) every nonzero homomorphism of rings R! S is a monomorphism.124

Def:A principal ideal ring which is an integral domain is called a principal ideal domain.125

Def:A nonzero element a of a commutative ring R is said to divide an element b 2 R (noted a j b) is there exists x 2 Rsuch that ax = b. Elements a; b of R are said to be associates if a j b and b j a.126

Thm: Let a; b and u be elements of a commutative ring R with identity.(i) a j b if and only if h b i � h a i;(ii) a and b are associates if and only if h a i = h b i;(iii) u is a unit if and only if u j r for all r 2 R;(iv) u is a unit if and only if hu i = R;(v) The relation \a is an associate of b" is an equivalence relation on R.(vi) If a = br with r 2 R a unit, then a and b are associates. If R is an integral domain, the converse is true.127

Def: Let R be a commutative ring with identity. An element c of R is irreducible provided that :(i) c is a nonzero nonunit; (ii) c = ab implies a or b is a unit.An element p of R is prime provided that:(i) p is a nonzero nonunit; (ii) p j ab implies p j a or p j b.128

Thm: Let p and c be nonzero elements in an integral domain R.(i) p is prime if and only if h p i is a nonzero prime ideal;(ii) c is irreducible if and only if h c i is maximal in the set S of all proper principal ideals of R;(iii) Every prime element of R is irreducible;(iv) If R is a principal ideal domain, then p is prime if and only if p is irreducible;(v) Every associate of an irreducible [prime] element of R is irreducible [prime].(vi) The only divisors of an irreducible element of R are its associates and the units of R.129

Derrick Stolee

2 RINGS Algebra Study Guide 2.4 Polynomial Rings

Def:An integral domain R is a unique factorization domain provided that:(i) every nonzero nonunit element a of R can be written a = c1c2 � � � cn with c1; : : : ; cn irreducible.(ii) If a = c1c2 � � � cn and a = d1d2 � � � dm (ci; di irreducible), then n = m and for some permutation � 2 Sn, ci andd�(i) are associates for every i.

130

Lma: If R is a principal ideal ring and h a1 i � h a2 i � � � � is a chain of ideals in R, then for some positive integer n,h aj i = h an i for all j � n.131

Thm: Every principal ideal domain R is a unique factorization domain.132

Def: Let N0 be the set of nonnegative integers and R a commutative ring. R is a Euclidean ring if there is a function' : R n f0g ! N0 such that(i) if a; b 2 R and ab 6= 0, then '(a) � '(ab);(ii) if a; b 2 R and b 6= 0, then there exist q; r 2 R such that a = qb+ r with r = 0, or r 6= 0 and '(r) < '(b).A Euclidean ring that is an integral domain is called a Euclidean domain.133

Ex: The Gaussian integers, Z[i], are a Euclidean domain. De�ne '(a+ bi) = a2 + b2.134

Thm: Every Euclidean ring R is a principal ideal ring with identity. Consequently every Euclidean domain is a uniquefactorization domain.135

Def: Let X be a nonempty subset of a commutative ring R. An element d 2 R is a greatest common divisor of Xprovided:(i) d j a for all a 2 X; (ii) c j a for all a 2 X implies c j d.If R has an identity and a1; : : : ; an have greatest common divisor 1R, then these elements are relatively prime.136

Thm: Let a1; : : : ; an be elements of a commutative ring R with identity.(i) d 2 R is a greatest common divisor of fa1; : : : ; ang such that d = r1a1 + � � �+ rnan for some ri 2 R if and only ifh d i = h a1 i+ � � �+ h an i;(ii) If R is a principal ideal ring, then a greatest common divisor of a1; : : : ; an exists and every one is of the formr1a1 + � � �+ rnan, ri 2 R;(iii) if R is a unique factorization domain, then there exists a greatest common divisor of a1; : : : ; an.

137

2.4 Polynomial Rings

Sections: Hungerford III.5,Thm: Let R be a ring and let R[x] denote the set of all sequences of elements of R(a0; a1; : : : ) such that ai = 0 for all

but a �nite number of indices i, called the ring of polynomials over R.(i) R[x] is a ring.(ii) if R is commutative [ring with identity, integral domain], then so is R[x].(iii) The map R! R[x] given by r 7! (r; 0; 0; : : : ) is a monomorphism of rings.138

Thm: Let R be a ring and denote by R[x1; : : : ; xn] the set of all functions f : Nn0 ! R such that f(u) 6= 0 for at most a�nite number of elements u 2 Nn0 . R[x1; : : : ; xn] is called the ring of polynomials in n indeterminants over R.(i) R[x1; : : : ; xn] is a ring.(ii) If R is commutative [ring with identity, has no zero divisors, integral domain], then so is R[x1; : : : ; xn].(iii) The map R ! R[x1; : : : ; xn] given by r 7! fr, where fr(0; 0; : : : ; 0) = r and f(u) = 0 otherwise, is a monomor-phism of rings.139

Thm: \Evaluation Principle" Let R and S be commutative rings with identity and ' : R! S a homomorphism of ringssuch that '(1R) = 1S . If s1; s2; : : : ; sn 2 S, then there is a unique homomorphism of rings ' : R[x1; : : : ; xn] ! Ssuch that 'jR = ' and '(xi) = si for i = 1; 2; : : : ; n. This property completely determines the polynomial ringR[x1; : : : ; xn] up to isomorphism.140

Cor: If ' : R! S is a homomorphism of commutative rings and s1; : : : ; sn 2 S, then the map R[x1; : : : ; xn]! S givenby f 7! 'f(s1; : : : ; sn) is a homomorphism of rings.141

Cor: Let R be a commutative ring with identity and n a positive integer. For each k(1 � k � n) there are isomorphismsof rings R[x1; : : : ; xk][xk+1; : : : ; xn] �= R[x1; : : : ; xn] �= R[xk+1; : : : ; xn][x1; : : : ; xk].

142

Prop: Let R be a ring and denote by R[[x]] the set of all sequences of elements of R(a0; a1; : : : ). This is called the ringof formal power series over R.(i) R[[x]] is a ring.(ii) The polynomial ring R[x] is a subring of R[[x]].(iii) If R is commutative [ring with identity, has no zero divisors, integral domain] then so is R[[x]].143

Prop: Let R be a ring with identity and f =P1i=0 aix

i 2 R[[x]].(i) f is a unit in R[[x]] if and only if its constant term a0 is a unit in R.

Derrick Stolee

2 RINGS Algebra Study Guide 2.5 Localizations

(ii) If a0 is irreducible in R, then f is irreducible in R[[x]].144

Cor: If R is a division ring, then the units in R[[x]] are precisely those power series with nonzero constant term. Theprincipal ideal hx i consists precisely of the nonunits in R[[x]] and is the unique maximal ideal of R[[x]]. Thus if Ris a �eld, R[[x]] is a local ring.145

Def: The degree of a monomial axk11 xk22 � � �xknn 2 R[x1; : : : ; xn] is the nonnegative integer k1 + k2 + � � � kn. The (total)

degree of the polynomial f 2 R[x1; : : : ; xn] is the maximum of the degrees of the monomials aixki;11 � � �x

ki;nn such that

ai 6= 0. A polynomial which is a sum of monomials, each of which has degree k, is homogeneous of degree k.146

Thm: Let R be a ring and f; g 2 R[x1; : : : ; xn].(i) deg(f + g) � maxfdeg f; deg gg. (ii) deg(fg) � deg f + deg g.(iii) If R has no zero divisors, deg(fg) = deg f + deg g.(iv) If n = 1 and the leading coe�cient of f or g is not a zero divisor in R (in particular, if it is a unit), thendeg(fg) = deg f + deg g.147

Thm: \Division Algorithm" Let R be a ring with identity and f; g 2 R[x] nonzero polynomials such that the leadingcoe�cient of g is a unit in R. Then there exist unique polynomials q; r 2 R[x] such that f = qg + r and deg r <deg g.148

Cor: \Remainder Theorem" Let R be a ring with identity and f(x) =Pni=0 aix

i 2 R[x]. For any c 2 R there exists aunique q(x) 2 R[x] such that f(x) = q(x)(x� c) + f(c).149

Cor: If F is a �eld, then the polynomial ring F [x] is a Euclidean domain, whence F [x] is a PID, UFD. The units inF [x] are precisely the nonzero constant polynomials.150

Def: Let R be a subring of a commutative ring S, c1; c2; : : : ; cn 2 S and f =Pni=0 aix

ki;11 � � �x

ki;nn 2 R[x1; : : : ; xn] a

polynomial such that f(c1; : : : ; cn) = 0. Then (c1; : : : ; cn) is a root or zero of f (or a solution of the polynomialequation f(x1; : : : ; xn) = 0).151

Thm: Let R be a commutative ring with identity and f 2 R[x]. Then c 2 R is a root of f if and only if x � c dividesf .152

Thm: If D is an integral domain contained in an integral domain E and f 2 D[x] has degree n, then f has at most ndistinct roots in E.153

Prop: \Rational Roots Theorem" Let D be a UFD with quotient �eld F and let f =Pni=0 aix

i 2 D[x]. If u = c=d 2 Fwith c and d relatively prime, and u is a root of f , then c divides a0 and d divides an.

154

Lma: Let D be an integral domain and f =Pni=0 aix

i 2 D[x]. Let f 0 2 D[x] be the polynomial f 0 =Pnk=1 kakx

k�1,the formal derivative of f . Then for all f; g 2 D[x] and c 2 D:(i) (cf)0 = cf 0; (ii) (f + g)0 = f 0 + g0; (iii) (fg)0 = f 0g + fg0; (gn)0 = ngn�1g0.155

Thm: Let D be an integral domain which is a subring of an integral domain E. Let f 2 D[x] and c 2 E.(i) c is a multiple root of f if and only if f(c) = 0 and f 0(c) = 0.(ii) If D is a �eld and f is relatively prime to f 0, then f has no multiple roots in E.(iii) If D is a �eld, f is irreducible in D[x] and E contains a root of f , then f has no multiple roots in E if and onlyif f 0 6= 0.156

Def: Let D be a UFD and let f =Pni=0 aix

i a nonzero polynomial in D[x]. A greatest common divisor of a0; a1; : : : ; anis called a content of f , denoted C(f). If a 2 D and f 2 D[x], then C(af) and aC(f) are the same up to associates.If f 2 D[x] and C(f) is a unit in D, then f is said to be primitive.157

Lma: \Gauss's Lemma" If D is a unique factorization domain and f; g 2 D[x], then C(fg) = C(f)C(g). In particular,the product of primitive polynomials is primitive.158

Lma: Let D be a UFD with quotient �eld F and let f and g be primitive polynomials in D[x]. THen f and g areassociates in D[x] if and only if they are associates in F [x].159

Lma: LetD be a UFD with quotient �eld F and f a primitive polynomial of positive degree inD[x]. Then f is irreduciblein D[x] if and only if f is irreducible in F [x].160

Thm: If D is a UFD, then so is the polynomial ring D[x1; : : : ; xn].161

Thm: \Eisenstein's Criterion" Let D be a UFD with quotient �eld F . If f =Pni=0 aix

i 2 D[x], deg f � 1 and p is anirreducible element of D such that p 6j an; p j ai for i = 0; 1; : : : ; n� 1; and p2 6j a0 then f is irreducible in F [x]. If fis primitive, then f is irreducible in D[x].162

2.5 Localizations

Sections: Hungerford III.4Def:A nonempty subset S � R, a ring, is multiplicative provided that a; b 2 S ) ab 2 S.163

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2 RINGS Algebra Study Guide 2.6 Dedekind Domains

Thm: Let S be mult-closed of commie ring R. The relation � on R� S given by (r; s) � (r0; s0), s1(rs0 � r0s) = 0 for

some s1 2 S is an equivalence relation.164

Thm: Let S be mult-closed of commie ring R and let S�1R be the equiv-classes of � above.165

(i) S�1R is a commie ring w/ identity.(ii) If R is a nonzero ring with no zero divisors and 0 =2 S, then S�1R is an integral domatin.(iii) If R is a nonzero ring with no zero divisors and S is the set of all nonzero elements of R, then S�1R is a �eld,called the �eld of fractions over R.

Thm: Let S be mult-closed of commie ring R.166

(i) 'S : R! S�1R given by r 7! rs=s is well-de�ned hom. of rings with 'S(s) is a unit in S�1R for all s 2 S.(ii) If 0 =2 S and S contains no zero divisors, then 'S is injective. In particular, any integral domain is embedded inits quotient �eld.(iii) If R has an identity and S consists of units, then 'S is an isomorphism. In particular, the complete ring ofquotients of a �eld is isomorphic to F .

Thm: Let S be mult-closed of commie ring R and let T be any commie ring with identity. If f : R ! T is a hom. ofrings such that f(s) is a unit in T for all s 2 S, then there exists a unique homomorphism of rings f : S�1R ! Tsuch that f � 'S = f . The ring S�1R is completely determined by this property.167

Cor: Let R be an integral domain considered as a subring of its quotient �eld F . If E is a �eld and f : R ! E aninjection of rings, then there is a unique injection of �eld f : F ! E so that f jR = f . In particular any �eld E1

containing R contains an isomorphic copy F1 of F with R � F1 � E1.168

Thm: Let S be mult-closed of a commie ring R.169 (i) If I is an ideal in R, then S�1I = fa=s j a 2 I; s 2 Sg is an idealin S�1R, called the extension of I.(ii) If J is another ideal in R, then(a) S�1(I + J) = S�1I + S�1J (b) S�1(IJ) = (S�1I)(S�1J) (c) S�1(I \ J) = S�1I \ S�1J

Thm: Let S be a mult-closed of commie ring R with identity and I and ideal of R. Then S�1I = S�1R if and only ifS \ I = ;.170

Lma: Let S be mult-closed of a commie ring R with identity and let I be an ideal of R.171

(i) I � '�1S (S�1I)(ii) If I = '�1S (J) for some ideal J in S�1R, then S�1I = J . In other words, every ideal in S�1R is of the formS�1I for some ideal I in R.(iii) If P is a prime ideal in R and S \ P = ;, then S�1P is a prime ideal in S�1R and ��1S (S�1P ) = P .

Thm: Let S be mult-closed of a commie ring R with identity. Then there is a one-to-one correspondence between theprime ideals of R disjoint from S and the set of prime ideals of S�1R given by P 7! S�1P .172

Def: Let R be a commie ring with identity and P a prime ideal of R. Then S = R � P is mult-closed and de�neRP = S�1R to be the localization of R at P .173

Def:A local ring is a commutative ring with identity which has a unique maximal ideal.174

Thm: Let P be a prime ideal in a commie ring R.175 (i) There is a one-to-one correspondence between the set of primeideals of R which are contained in P and the set of prime ideals of RP , given by Q 7! QP .(ii) The ideal PP in RP is the unique maximal ideal of RP . Hence, RP is local.

Thm: If R is a commutative ring with identity, then the following conditions are equivalent.(i) R is a local ring;(ii) all nonunits of R are contained in some ideal M 6= R;(iii) the nonunits or R form an ideal.176

2.6 Dedekind Domains

Def:A Dedekind domain is an integral domain R in which every ideal (6= R) is the produce of a �nite number ofprime ideals.177

Def: Let R be an integral domain with quotient �eld K. A fractional ideal of R is a nonzero R-submodule I of K suchthat aI � R for some nonzero a 2 R.178

Thm: If R is an integral domain with quotient �eld K, then the set of all fractional ideals of R forms a commutativemonoid, with identity R and multiplication given by IJ = f

Paibi j ai 2 I; bi 2 JU; n 2 Ng.179

Def:A fractional ideal I of an integral domain R is said to be invertible if IJ = R for some fractional ideal J of R.180

Lma: Let I; I1; : : : ; In be ideals in an integral domain R.181

(i) The ideal I1I2 � � � In is invertible if and only if every Ij is invertible.

Derrick Stolee

NOTES Algebra Study Guide NOTES

(ii) If P1 � � �Pm = I = Q1 � � �Qn where each Pi; Qj is prime in R and Pi is invertible, then m = n and Pi = Qi witha suitable reordering.

Thm: If R is a Dedekind domain, then every nonzero prime ideal of R is invertible and maximal.182

Lma: If I is a fractional ideal of an integral domain R with quotient �eld K and f 2 HomR(I;R) then for all a; binI,af(b) = bf(a).183

Lma: Every invertible fractional ideal of an integral domain R with quotient �eld K is a �nitely generated R-module.184

Thm: Let R be an integral domain and I a fractional ideal of R. Then I is invertible if and only if I is a projectiveR-module.185

Lma: If R is a Noetherian, integrally closed integral domain and R has a unique nonzero prime ideal P , then R is adiscrete valuation ring.186

Thm: The following conditions on an integral domain R are equivalent:187 (i) R is a Dedekind domain.(ii) Every proper ideal in R is uniquely a product of a �nite number of prime ideals.(iii) Every nonzero ideal in R is invertible.(iv) Every fractional ideal of R is invertible.(v) The set of all fractional ideals of R is a group under multiplication.(vi) Every ideal in R is projective.(vii) Every fractional ideal of R is projective.(viii) R is Noetherian, integrally closed, and every nonzero prime ideal is maximal.(ix) R is Notherian and for every nonzero prime ideal P of R, the localization RP of R at P is a discrete valuationring.

Artin: Algebra by Michael Artin.D&F: Abstract Algebra by Dummit & Foote.H: Algebra by Thomas W. Hungerford.Lang: Algebra by Serge Lang.

Notes

79H Def 1.1 p.11580H Thm 1.2 p.11581H Def 1.3 p.11682H Def 1.4 p.11683H Thm 1.6 p.11884H Def 1.7 p.11885H pp.118-986H Def 1.8 p.11987H Thm 1.9 p.11988H Thm 1.10 p.11989H Ex 12 p.12190H Def 2.1 p.12291H Eg p.12292H Eg/Rmk p.12393H Thm 2.2 p.12394H Cor 2.3 p.12395H Def 2.4 p.12396H Thm 2.5 pp.123-497H Thm 2.6 p.12498H Thm 2.7 p.12599H Thm 2.8 p.125100H Thm 2.9 p.125101H Cor 2.10 p.126102H Cor 2.11 p.126103H Thm 2.12 p.126104H Them 2.13 p.126105H Def 2.14 p.126106H Thm 2.15 p.127107H Def 2.17 p.127108Class 01/18109H Thm 2.18 p.128110H Thm 2.19

111H Rmk p.128112H Thm 2.22 pp.129-30113H Thm 2.23 p.130114H Thm 2.24 p.130115H p.131116H Thm 2.25 p.131117H Cor 2.26 p.132118H Cor 2.27 p.132119H Exs 23,24 p.135120H Def 1.5 p.116121H Eg p.127122H Thm 2.16 p.127123H Thm 2.20124H Cor 2.21 p.129125H Def 2.4 p.123126H Def 3.1 p.135127H Thm 3.2 p.136128H Def 3.3 p.136129H Thm 3.4 p.136130H Def 3.5 p.137131H Lma 3.6 p.137132H Thm 3.7133H Def 3.8 p.139134H Eg p.139135H Thm 3.9 p.139136H Def 3.10 p.140137H Thm 3.11 p.140138H Thm 5.1 p.149139H Thm 5.3 p.151140H Thm 5.5 p.152141H Cor 5.6 p.153142H Cor 5.7 p.153143H Prop 5.8 p.154144H Prop 5.9 p.155145H Cor 5.10 p.155146H pp.157-8147H Thm 6.1 p.158148H Thm 6.2 p.158149H Cor 6.3 p.159150H Cor 6.4 p.159151H Def 6.5 p.160

Derrick Stolee

NOTES Algebra Study Guide NOTES

152H Thm 6.6 p.160153H Thm 6.7 p.160154H Prop 6.8 p.161155H Lma 6.9 p.161156H Thm 6.10 p.161157H p.162158H Lma 6.11 p.162159H Lma 6.12 p.163160H Lma 6.13 p. 163161H Thm 6.14 p.164162H Thm 6.15 pp.164-5163H Def 4.1 p.142164H Thm 4.2 p.142165H Thm 4.3 p.143166H Thm 4.4 p.144167H Thm 4.5 p.144168H Cor 4.6 p.145169H Thm 4.7 p.145

170H Thm 4.8 p.146171H Lma 4.9 p.146172H Thm 4.10 p.146173H p.147174H Def 4.12 p.147175H Thm 4.11 p.147176H Thm 4.13 p.147177H p.401178H Def 6.2 p.401179H Thm 6.3 p.401180H p.401181H Lma 6.4 p.402182H Thm 6.5 p.402183H Lma 6.6 p.403184H Lma 6.7 p.403185H Thm 6.8 p.404186H Lma 6.9 p.404187H Thm 6.10 pp.405-6

Derrick Stolee

3 MODULES Algebra Study Guide

3 Modules

3.1 De�nitions

Def: Let R be a ring. A [left] R-module is an additive abelian group A together with a function R�A! A (writtenas multiplication) such that for all r; s 2 R and a; b 2 A:(i) r(a+ b) = ra+ rb. (ii) (r + s)a = ra+ sa. (iii) r(sa) = (rs)a.If R has an identiy element 1R, and(iv) 1Ra = a for all a 2 A,then A is a unitary R-module. If R is a division ring, then a unitary R-module is called a [left] vector space.188

Ex: If R is a ring, every abelian group A can be made into an R-module with trivial module structure by de�ning ra = 0for all r 2 R and a 2 A.H Eg p.170

Def:A non-zero unitary R-module A is simple if its only submodules are 0 and A.189

Def: Let A and B be modules over a ring R. A function f : A ! B is an R-module homomorphism provided for alla; c 2 A and r 2 R, f(ra + c) = rf(a) + f(c). If R is a division ring, then an R-module homomorphism is called alinear transformation. De�ne similarly monomorphism, epimorphism, isomorphism, image and kernel.190

Def: Let R be a ring, A and R-module and B a nonempty subset of A. B is a submodule of A provided B is an additivesubgroup of A and rb 2 B for all r 2 R, b 2 B. A submodule of a vector space is called a subspace.191

Def: If X is a subset of a module A over a ring R, then the intersection of all submodules of A containing X is calledthe submodule generated by X (or spanned by X). If X is �nite and X generates the module B, then B is said tobe �nitely generated. If X consists of a single element X = fag, then the submodule generated by X is called thecyclic (sub)module generated by a. If fBi j i 2 Ig is a family of submodules of A, then the submodule generated byX = [i2IBi is called the sum of the modules Bi. If I is �nite, label this as B1 +B2 + � � �+Bn.

192

Thm: Let R be a ring, A and R-mod, X � A, fBi j i 2 Ig a family of submodules of A and a 2 A. Let Ra = fra j r 2 Rg.(i) Ra is a submodule of A and the map R! Ra given by r 7! ra is an R-mod epimorphism.(ii) The cyclic submodule C generated by a is fra+ na j r 2 R;n 2 Zg. If R has an identity and C is unitary, thenC = Ra.(iii) The submodule D generated by X is

nPsi=1 riai +

Ptj=1 njbj j s; t 2 N; ai; bj 2 X; ri 2 R;ni 2 Z

o.

If R has an identity and A is unitary, then D = RX = fPsi=1 riai j s 2 N; ai 2 X; ri 2 Rg.

(iv) The sum of the family fBi j i 2 Ig consists of all �nite sums bi1 + � � �+ bin with bik 2 Bik .193

Thm: Let B be a submodule of a module A over a ring R. Then the quotient group A=B is an R-module with the actionof R on A=B given by r(a + B) = ra + B for all r 2 R; a 2 A. The map � : A ! A=B given by a 7! a + B is anR-mod epimorphism with kernel B, called the canonical epimorphism or projection.194

Thm: \First Isomorphism Theorem of Modules" If R is a ring and f : A! B is an R-module homomorphism and C isa submodule of ker f , then there is a unique R-mod homomorphism f : A=C ! B such that f(a+C) = f(a) for alla 2 A, im f = im f , and ker f = (ker f)=C. f is an R-mod isomorphism if and only if f is an R-module epimorphismand C = ker f . In particular, A= ker f �= im f .195

Cor: If R is a ring and A0 is a submodule of the R-module A and B0 a submodule of the R-module B and f : A! B isan R-module homomorphism such that f(A0) � B0, then f induces an R-module homomorphism f : A=A0 ! B=B0

given by a + A0 7! f(a) + B0. f is an R-module isomorphism if and only if im f + B0 = B and f�1(B0) � A0. Inparticular if f is an epimorphism such that f(A0) = B0 and ker f � A0,t hen f is an R-module isomorphism.196

Thm: Let B and C be submodules of a module A over a ring R.(i) \Second Isomorphism Theorem of Modules"There is an R-module isomorphism B=(B \ C) �= (B + C)=C;(ii) \Third Isomorphism Theorem of Modules" if C � B, then B=C is a submodule of A=C, and there is an R-moduleisomorphism (A=C)=(B=C) �= A=B.197

Thm: If R is a ring and B a submodule of an R-mod A, then there is a one-to-one correspondence between the set of allsubmodules of A containing B and the set of all submodules of A=B given by C 7! C=B. Hence every submodule ofA=B is of the form C=B, where C is a submodule of A which contains B.198

Thm: Let R be a ring and fAi j i 2 Ig a nonempty family of R-modules,Qi2I Ai the direct product of the abelian

groups Ai, andPi2I Ai the direct sum of the abelian groups Ai. If I is �nite, then

QAi =

PAi = A1 � � � � �An.

(i)Qi2I Ai is an R-module with the action of R given by rfaig = fraig.

(ii)Pi2I Ai is a submodule of

Qi2I Ai.

(iii) For each k 2 I, the canonical projection �k :QAi ! Ak is an R-module epimorphism.

(iv) For each k 2 I, the canonical injection {k : Ak !PAi is an R-module monomorphism.199

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3 MODULES Algebra Study Guide 3.2 Free and Projective Modules

Thm: If R is a ring, fAi j i 2 Ig a family o� R-modules, C an R-mod, and f'i : C ! Ai j i 2 Ig a family of R-modulehomomorphisms, then there is a unique R-mod homomorphism ' : C !

QAi such that �i' = 'i for all i 2 I.

QAi

is uniquely determined up to isomorphism by this property, soQAi is a product in the category of R-modules.200

Thm: If R is a ring, fAi j i 2 Ig a family of R-modules, D an R-module, and f i : Ai ! D j i 2 Ig a family ofR-module homomorphisms, then there is a unique R-module homomorphism :

PAi ! D such that {i = i for

all i 2 I.PAi is uniquely determined up to isomorphism by this property, so

PAi is a coproduct in the category

of R-modules.201

Thm: Let R be a ring and A;A1; : : : ; An R-modules. Then A �= A1�� � ��An if and only if for each i = 1; 2; : : : ; n thereare R-module homomorphisms �i : A! Ai and { : Ai ! A such that(i) �i{i = idAi

for all i;(ii) �j{i = 0 for i 6= j;(iii) {1�1 + � � �+ {n�n = 1A.

202

Thm: Let R be a ring and fAi j i 2 Ig a family of submodules of an R-mod A such that(i) A is the sum of the family fAi j i 2 Ig;(ii) for each k 2 I, Ak \A

�k = 0, where A�k is the sum of the family fAi j i 6= kg.

Then there is an isomorphism A �=PAi and A is called the (internal) direct sum of fAi j i 2 Ig.

203

Def:A pari of module homomorphisms, Af! B

g! C is exact at B if im f = ker g. A �nite sequence of module

homomorphisms, A0f1! A1

f2! � � �

fnto An is exact provided im fi = ker fi+1 for i = 1; : : : ; n�1. An in�nite sequence of

module homomorphisms is exact provided im fi = ker fi+1 for all i 2 Z. A �nite exact sequence starting and endingwith the trivial module is a short exact sequence.204

3.2 Free and Projective Modules

Def:A subset X of an R-mod A is linearly independent provided that for distinct x1; : : : ; xn 2 X and ri 2 R,Pni=1 rixi = 0 implies ri = 0 for all i. A set that is not linearly independent is linearly dependent. If A is generated

by linear combinations of the elements of X, then X spans A. A linearly independent subset of A that spans A is abasis of A. An R-mod F that has a nonempty basis is called a free R-module.205

Thm: Let R be a ring with identity. The following conditions on a unitary R-module F are equivalent:(i) F has a nonempty basis;(ii) F is the internal direct sum of a family of cyclic R-modules, each of which is isomorphic as a left R-module to R;(iii) F is R-module isomorphic to a direct sum of copies of the left R-module R; (iv) there exists a nonempty setX and a function { : X ! F with the following property: given any unitary R-module A and function f : X ! A,there exists a unique R-module homomorphism f : F ! A such that f{ = f . In other words, F is a free object inthe category of unitary R-modules.206

Cor: Every [unitary] module A over a ring R [with identity] is the homomorphic image of a free R-module F . If A is�nitely generated, then F may be chose to be �nitely generated.207

Lma:A maximal linearly independent subset X of a vector space V over a division ring D is a basis of V .208

Thm: Every vector space V over a division ring D has a basis and is therefore a free D-module. More generally everylinearly independent subset of V is contained in a basis of V .209

Thm: If V is a vector space over a division ring D and X is a subset that spans V , then X contains a basis of V .210

Thm: Let R be a ring with identity and F a free R-module with an in�nite basis X. Then every basis of F has thesame cardinality as X.211

Thm: If V is a vector space over a division ring D, then any two bases of V have the same cardinality.212

Def: Let R be a ring with identity such that for every free R-module F , any who bases of F have the same cardinality.THen R is said to have the invariant dimension property and the cardinal number of any basis of F is called thedimension (or rank) of F over R.213 A vector space V over a division ring D is �nite dimensional if dimD V is�nite.214

Prop: Let E and F be free modules over a ring R that has the invariant dimension property. Then E �= F if and onlyif E and F have the same rank.215

Lma: Let R be a ring with identity, I( 6= R) an ideal of R, F a free R-module with basis X and � : F ! F=IF thecanonical epimorphism. Then F=IF is a free R=I-module with basis �(X) and j�(X)j = jXj.216

Prop: Let f : R! S be a nonzero epimorphism of rings with identity. If S has the invariant dimension property, thenso does R.217

Derrick Stolee

3 MODULES Algebra Study Guide 3.3 Modules over PIDs

Cor: If R is a ring with identity that has a homomorphic image which is a division ring, then R has the invariantdimension property. In particular, every commutative ring with identity has the invariant dimension property.218

Thm: Let W be a subspace of a vector space V over a division ring D.(i) dimDW � dimD V ; (ii) if dimDW = dimD V and dimD V is �nite, then W = V ;(iii) dimD V = dimDW + dimD(V=W ):219

Cor: If f : V ! V 0 is a linear transformation of vector spaces over a division ring D, then there exists a basis Xof V such that X \ ker f is a basis of ker f and ff(x) j f(x) 6= 0; x 2 Xg is a basis of im f . In particular,dimD V = dimD(ker f) + dimD(Im f).220

Cor: If V and W are �nite dimensional subspaces of a vector space over a division ring D, then dimD V + dimDW =dimD(V \W ) + dimD(V +W ).221s

Thm: Let R;S; T be division rings such that R � S � T . Then dimR T = (dimS T )(dimR S): Furthermore, dimR T is�nite if and only if dimS T and dimR S are �nite.222

Def:A module P over a ring R is projective if given any R-hom f : P ! B and exact sequence Ag! B ! 0, there

exists a map h : P ! A so that g � h = f .223

Thm: Every free module F over a ring R with identity is projective.224

Cor: Every module A over a ring R is the homomorphis image of a projective R-module.225

Thm: Let R be a ring. The following conditions on an R-mod P are equivalent:226 (i) P is projective.

(ii) Every short exact sequence 0! Af! B

g! P ! 0 is split exact, hence B �= A� P .

(iii) there is a free module F and an R-mod K so that F �= K � P .

Prop: Let R be a ring. A direct sum of R-modsX

Pi is projective if and only if each Pi is projective.227

3.3 Modules over PIDs

3.4 Exact Sequences and Homs

Thm: Let A;B;C;D be module over a ring R and ' : C ! A, : B ! D are R-homs. Then the map � :HomR(A;B)! HomR(C;D) given by f 7! f' is a homomorphism of abelian groups.228

Thm: Let R be a ring. 0 ! A'! B

! C is an exact sequence or R-modules if and only if for every R-module D,

0! HomR(D;A)'! HomR(D;B)

! HomR(D;C) is an exact sequence of abelian groups.229

Prop: Let R be a ring. A'! B

! C ! 0 is an exact sequence or R-modules if and only if for every R-module D,

0! HomR(C;D) ! HomR(B;D)

'! HomR(A;D) is an exact sequence of abelian groups.230

Def: The theorems above show that HomR(A;B) is left exact.231

Prop: The following conditions on modules over a ring R are equivalent:232

(i) 0! A'! B

! C ! 0 is a split exact sequence of R-modules.

(ii) 0! HomR(D;A)'! HomR(D;B)

! HomR(D;C) is an exact sequence of abelian groups for every R-mod D.

(iii) 0! HomR(C;D) ! HomR(B;D)

'! HomR(A;D) is an exact sequence of abelian groups for every R-mod D.

Thm: The following conditions on a module P over a ring R are equivalent.233

(i) P is projective.(ii) If : B ! C is any injective R-hom then : HomR(P;B) ! HomR(P;C) is an injective homomorphism ofabelian groups.

(iii) If 0 ! A'! B

! C ! 0 is any short exact sequence of R-mods, then 0 ! HomR(P;A)

'! HomR(P;B)

!

HomR(P;C)! 0 is an exact sequence of abelian groups.

Thm: LetA;B; fAi j i 2 Ig; fBj j j 2 Jg beR-mods. Then there are isomorphism of abelian groups: (i) HomR

Xi2I

Ai; B

!�=Y

i2I

HomR(Ai; B)

(ii) HomR

0@A;Y

j2J

Bj

1A �=

Yj2J

HomR(A;Bj).234

Def:An abelian group A is an R � S bimodule provided that A is both a left R-module and a right S-module andr(as) = (ra)s for all r 2 R; s 2 S.235

Derrick Stolee

3 MODULES Algebra Study Guide 3.5 Tensor Products

Thm: Let R and S be rings and RA;RBS ;R CS ;RD be (bi)modules as indicated.236

(i) HomR(A;B) is a right S-module, with the action of S given by (fs)(a) = (f(a))s; f 2 HomR(A;B); s 2 S; a 2 A.(ii) If ' : A! A0 is a left R-hom, then the induced map ' : HomR(A

0; B)! HomR(A;B) is a right S-hom.(iii) HomR(C;D) is a left S-mod with the action of S given by (sg)(c) = g(cs) for g 2 HomR(C;D); s 2 S; c 2 C.

Thm: If A is a unitary left R-mod, then there is an isomorphism of left R-modules A �= HomR(R;A).237

Def:Given A a left R-module, consider R as a right R-mod and de�ne A� = HomR(A;R), the dual module of A. Theelements of A� are sometimes called linear functionals.238

Thm: Let A;B and C be left R-mods.239 (i) If ' : A! C os a R-hom, then the induced map ' : C� = HomR(C;R)!HomR(A;R) = A� is a right R-hom.(ii) There is an R-mod isomorphism (A� C)� �= A� � C�.(iii) If R is a division ring, and 0 ! A ! B ! C ! 0 is a short exact sequence of left vector spaces, then0! C� ! B� ! A� ! 0 is a short exact sequence of right vector spaces.

Def: The Kronecker delta notation is the symbol �ij =

(0R i 6= j

1R i = j:240

Thm: Let F be a free left R-0mod. Let X be a basis of F and for each x 2 X let fx : F ! R be de�ned by the operationon the basis fx(y) = �xy. Then,(i) ffx j x 2 Xg is a linearly independent subset of F � of cardinality jXj;(ii) if X is �nite, then F � is a free right R-module with basis ffx j x 2 Xg, called the dual basis.241

3.5 Tensor Products

Def:A middle linear map from AR�R B to C is a function f : A�B ! C such that for all a; ai 2 A; b; bi 2 B; andr 2 R,(i) f(a1 + a2; b) = f(a1; b) + f(a2; b) (ii) f(a; b1 + b2) = f(a; b1) + f(a; b2) (iii) f(ar; b) = f(a; rb).242

Def: Let A be a right R-mod and B a left R-mod. Let F be the free abelian group on the set A � B. Let K be thesubgroup of F generated by all elements of the forms:(i) (a+ a0; b)� (a; b)� (a0; b) (ii) (a; b+ b0)� (a; b)� (a; b0) (iii) (ar; b)� (a; rb).The quotient group F=K is called the tensor product of A and B, denoted AR B. The coset (a; b) +K is denoteda b.243

Thm: Let AR and RB be R-mods and let C be an abelian group. If g : A� B ! C is a middle linear map, then thereexists a unique group hom g : AR B ! C so that g{ = g where { : A�B ! AR B is the canonical middle linearmap. ARB is uniquely determined up to isomorphism y this property. Therefore, { : A�B ! ARB is universalin the category of all middle linear maps on A�B.244

Cor: If AR; A0R;RB and RB

0 are R-mods and f : A ! A0 and g : B ! B0 are R-homs, then there is a unique grouphom AR B ! A0 R B

0 such that a b 7! f(a) g(b) for all a 2 A; b 2 B.245

Prop: If A ! B ! C ! 0 is an exact sequence of left R-mods and D is a right R-mod, then D R A ! D R B !DRC ! 0 is an exact sequence of abelian groups. An analogous statement hods for exactness in the �rst position.246

Thm: Let R and S be rings and SAR;RB;CR;RDS (bi)mods as indicated.247

(i) AR B is a left S-mod so that s(a b) = sa b for all s 2 S; a 2 A.(ii) If f : A! A0 is an S �R-hom and g : B ! B0 is an R-hom, then the induced map f g : AR B ! A0 R B

0

is a left S-hom.(iii) If h : C ! C 0 is an R-hom and k : D ! D0 is an R�S-hom, then the induced map h k : C RD ! C 0RD

0

is a right S-hom.Thm: If A;B;C are modules over a commutative ring R and g : A � B ! C is a bilinear map, then there is a unique

R-hom g : AR B ! C so that g{ = g. AR B is uniquely determined up to isomorphism by this property.248

Thm: If R is a ring with identity and AR;RB are unitary R-mods, then there are R-mod isomorphisms A R R �= Aand RR B �= B.249

Thm: If R and S are rings and AR;RBS ;S C are (bi)modules, then there is an isomorphism (A R B) S C �= A R(B S C): The isomorphism is unique given the restriction (a b) c 7! a (b c).250

Prop: Let E;F be modules of a commie ring R. Then there is a unique isomorphism E R F ! F R E so thate f 7! f e.251

Thm: Let R be a ring, A and fAi j i 2 Ig right R-mods, B and fBj j j 2 Jg left R-modes. Then there are groupisomorphisms

Derrick Stolee

NOTES Algebra Study Guide NOTES

(i)�X

Ai

�R B �=

X(Ai R B);

(ii) AR�X

Bj

��=X

(AR Bj).252

Thm: \Adjoint Associativity" Let R and S be rings and AR;RBS ; CS (bi)modules. Then there is an isomorphism ofabelian groups � : HomS(AR B;C) �= HomR(A;HomS(B;C)), de�ned for each f : AR B ! C by [(�f)(a)](b) =f(a b).253

Thm: Let R be a ring with identity. If A is a unitary right R-mod and V is a free left R-mod with basis Y , then every

element u of AR F may be written uniquely in the form u =

nXi=1

ai yi where ai 2 A and yi are distinct elements

of Y .254

Cor: If R is a ring with identity and AR and RB are free R-modules with bases X and Y , then ARB is a free (right)R-mod with basis W = fx y j x 2 X; y 2 Y g of cardinality jXj � jY j.255

Cor: Let S be a ring with identity and R a subring of S that contains 1S . If F is a free left R-mod with basis X, thenS R F is a free left S-module with basis f1S x j x 2 Xg.256

Prop: Let E;F be free commie R-mods of �nite dimension. Then we have an isomorphism EndR(E) EndR(F ) !EndR(E F ) which is the unique maps such that f g 7! T (f; g) for f 2 EndR(E); g 2 EndR(F ).

257

Def: Let E;F;G be modules of a commie ring R. Then L(E;F ) is the set of linear maps E ! F . The set L2(E;F ;G)is the set of bilinear maps E � F ! G.258

Prop: For a commie ring R and three R-mods E;F; and G, L(E;L(F;G)) �= L2(E;F ;G) �= L(E F;G).259

Prop: Let 0! E0 ! E ! E00 ! 0 be an exact sequence, and F andy module. Then the sequence F E0 ! F E !F E00 ! 0 is exact. Therefore, the tensor product is right exact.260

Prop: Let I be an ideal of a commie ring R. Let E be an R-mod. Then the map (R=I) � E ! E=IE induced by(a; x) 7! ax (mod IE); a 2 R; x 2 E is bilinear and induces an isomorphism (R=I)R E �= E=IE.261

3.5.1 Flat Modules

Artin: Algebra by Michael Artin.D&F: Abstract Algebra by Dummit & Foote.H: Algebra by Thomas W. Hungerford.Lang: Algebra by Serge Lang.

Notes

188H Def 1.1 p.169189H Ex 5 p.179190H Def 1.2 p.170191H Def 1.3 p.171192H Def 1.4 p.171193H Thm 1.5 pp.171-2194H Thm 1.6 p.172195H Thm 1.7 p.172196H Cor 1.8 p.172197H Thm 1.9 p.173198H Thm 1.10 p.173199H Thm 1.11 p.173200H Thm 1.12 p.173201H Thm 1.13 p.174202H Thm 1.14 p.174203H Thm 1.15 p.175204H Def 1.16 pp.175-6205H p.181206H Thm 2.1 p.181207H Cor 2.2 p.182208H Lma 2.3 p.183209H THm 2.4 p.183210H Thm 2.5 p.183211H Thm 2.6 p.184212H Thm 2.7 p.185

213H Def 2.8 p.185214H p.186215H Prop 2.9 p.185216H Lma 2.10 p.185217H Prop 2.11 p.186218H Cor 2.12 p.186219H Thm 2.13 p.187220H Cor 2.14 p.187221H Cor 2.15 p.187222H Thm 2.16 p.188223H Def 3.1 pp.190-1224H Thm 3.2 p.191225H Cor 3.3 p.192226H Thm 3.4 p.192227H Prop 3.5 p.193228H Thm 4.1 p.199229H Thm 4.2 p.200230H Prop 4.3 p.200231H p.201232H Prop 4.4 p.201233H Thm 4.5 p.201234H Thm 4.7 p.202235H p.202236H Thm 4.8 p.203237H Thm 4.9 p.203238H p.203239H Thm 4.10 p.204240H p.204241H Thm 4.11 p.204242H p.207243H Def 5.1 p.208; L p.602244H Thm 5.2 p.209245H Cor 5.3 p.209246H Prop 5.4 pp.209-210

Derrick Stolee

NOTES Algebra Study Guide NOTES

247H Thm 5.5 p.210248H Thm 5.6 p.211249H Thm 5.7 p.212250H Thm 5.8 p.212; L Prop 1.1 p.604251L Prop 1.2 p.605252H Thm 5.9 p.213; L Prop 2.1, Cor 2.2 p.608253H Thm 5.10 p.214254H Thm 5.11; L Prop 2.3 p.609

255H Cor 5.12 p.215; L Cor 2.4 p.609256H Cor 5.12257L Prop 2.5 p.610258L p.607259L p.607260L Prop 2.6 p.611261L Prop 2.7 p.612

Derrick Stolee

4 FIELD THEORY Algebra Study Guide

4 Field Theory

4.1 Basic Extensions

Def:A �eld F is said to be an extension �eld of k provided that k is a sub�eld of F .262

Def:A �eld extension k � F operates as a vector space. The dimension of this vector space is written as [F : k] = dimk Fand called the degree of the �eld extension.263

Def: Let F=k be a �eld extension. An element u 2 F is algebraic over k if u is a root of some polynomial in k[x].Otherwise, u is transcendental over k. F is called an algebraic extension if every element of F is algebraic over k.Otherwise, F is called a transcendental extention if at least one element of F is transcendental over k.264

Prop: Let F be a �nite extension over k, then F=k is algebraic.265

Def:A �eld E is called an intermediate �eld of F=k if k � E � F are �eld extensions.266

Thm: Let F be an extension �eld of E and E and extension �eld of k. Then [F : k] = [F : E][E : k]. Furthermore,[F : k] is �nite if and only if [F : E] and [E : k] are �nite.267 Also, if fxigi2I is a basis for E over k and fyjgj2J is abasis for F over E, then fxiyjg(i;j)2I�J is a basis for F over k.268

Def:A tower of �elds is a sequence k � F1 � F2 � � � � � Fn. A tower is �nite if every step of the sequence is �nite.269

Def: The �eld k(u1; : : : ; un) is called a �nitely generated extension of k. An extension k(u) is called simple.270

Prop: Let F be a �nite extension over k. Then F is �nitely generated.271

Thm: If F=k a �eld extension, u; ui 2 F and X � F , then(i) the subring k[u] consists of all elements of the form f(u) where f 2 k[x].(ii) the subring k[u1; : : : ; um] consists of all elements of the form g(u1; : : : ; um) where g 2 k[x1; : : : ; xm].(iii) The subring k[X] consists of all elements of the form h(u1; : : : ; um) where h 2 k[x1; : : : ; xm] and each ui 2 X,for some m.(iv) The sub�eld k(u) consists of all elements of the form f(u)=g(u) = f(u)g(u)�1 where f; g 2 k[x] and g(u) 6= 0.(v) The sub�eld k(u1; : : : ; um) consists of all elements of the form f(u1; : : : ; um)=g(u1; : : : ; um) where f; g 2 k[x1; : : : ; xm]and g(u1; : : : ; um) 6= 0.(vi) The sub�eld k(X) consists of all elements of the form f(u1; : : : ; um)=g(u1; : : : ; um) where ui 2 X for some m,f; g 2 k[x1; : : : ; xm] and g(u1; : : : ; um) 6= 0.272

Def: If L and M are sub�elds of a �eld F , the compositum (or sometimes composite) of L and M in F , denoted LM isthe sub�eld generated by the set L [M .273 Given two extensions E and F of a base �eld k, the extension EF overF as the translation or lifting of E to F .274

Def:K(x1; : : : ; xn) is called the �eld of rational functions in x1; : : : ; xn over k.275

Thm: If F=k a �eld extension, and u 2 F transcendental over k, then there is an isomorphism of �elds k(u) �= k(x)which is the identity on k.276

Thm: If F=k a �eld extension, u; ui 2 F algebraic over k, then(i) k(u) = k[u].(ii) k(u) �= k[x]= h f i, where f 2 k[x] is an irreducible monic polynomial of degree n � 1 uniquely determined by theconditions that f(u) = 0 and g(u) = 0(g 2 k[x]) if and only if f divides g.277

(iii) [k(u) : k] = deg f = n.278

(iv) f1k; u; u2; : : : ; un�1g is a basis for k(u) over k.279 (v) k(u1; : : : ; un) = k[u1; : : : ; un].

(vi) k[u1; : : : ; un]=k algebraic.280

Def: Let F=k a �eld extension and u 2 F algebraic over k. The monic irreducible polynomial such that f(u) = 0 iscalled the irreducible polynomial of u, written f = IrrPolyk(u). The degree of u over k is deg f = [k(u) : k].281

Thm: Let � : k ! L be an isomorphism of �elds, u and element of some extension �eld of k and v an element of someextension �eld of L. Assume either(i) u is transcendental over k and v is transcendental over L; or(ii) u is a root of an irreducible polynomial f 2 k[x] and v is a root of �(f) 2 L[x].Then � extends to an isomorphism of �elds k(u) �= L(u) which maps u onto v.282

Cor: Let E and F each be extensions of k and let u 2 E and v 2 F be algebraic over k. Then u and v are roots of thesame irreducible polynomial f 2 k[x] if and only if there is an isomorphism of �elds k[u] �= k[v] which sends u ontov and is the identity of k.283

Thm: If k is a �eld and f 2 k[x] a polynomial of degree n, then there exists a simple extension �eld F = k[u] of k suchthat:(i) u 2 F is a root of f .

Derrick Stolee

4 FIELD THEORY Algebra Study Guide 4.2 Normal Extensions

(ii) [k[u] : k] � n with equality holding if and only if f is irreducible in k[x].(iii) If f is irreducible in k[x], then k[u] is unique up to an isomorphism which is the identity on k.284

Thm: If F is a �nite dimensional extension �eld of k, then F is �nitely generated and algebraic over k.285

Thm: If F=k a �eld extension and X is a subset of F such that F = k(X) and every element of X is algebraic over k,then F is an algebraic extension of k. If X is a �nite set, then F is �nite dimensional over k.286

Thm: If k � E � F with E=k algebraic and F=E algebraic, then F=k is algebraic.287

Thm: Let F=k be a �eld extension and E the set of all elements of F algebraic over k. Then E is a sub�eld of F ,algebraic over k.288

4.1.1 Distinguished Classes of Field Extensions

Def: Let C be a certain class of extension �eld F � E. C is distinguished if it satis�es the following conidtions:(i) Let k � F � E a tower of �elds. The extension k � E is in C if and only if k � F and F � E are in C.(ii) If k � E is in C and F any extension of k, and E;F are both contained in some �eld, then F � EF is in C.(iii) If k � F and k � E are in C and F;E are sub�elds of a common �eld, then k � EF is in C.289

Prop: The class of algebraic extensions is distinguished, and so is the class of �nite extensions.290

Thm: Separable extensions form a distinguished class of extensions.291

4.2 Normal Extensions

Def: Let k be a �eld and f 2 k[x] with deg f � 1. An extension F=k is a splitting �eld for f if f splits into linearfactors in F [x].292 For a family ffigi2I � k[x], the splitting �eld of this family is an extension F=k such that eachfi splits in F .

293

Thm: Let K be a splitting �eld of the polynomial f(x) 2 k[x]. If E is another splitting �eld of f , then there exists anisomorphism � : E ! K inducing the identity on k. If k � K � ka, where ka is an algebraic closure of k, then anyembedding of E in ka inducing the identity on k must be an isomorphism of E onto K.294

Cor: Let K be a splitting �eld for the family ffigi2I and let E be another splitting �eld. Any embedding of E into Ka

inducing the identity on k gives an isomorphism of E onto K.295

Thm: \Conditions for Normal Extensions" Let K be an algebraic extension of k, contained in an algebraic closure ka ofk. THen the following conditions are equivalent:(i) Every embedding of K in ka over k induces an automorphism of K.(ii) K is the splitting �eld of a family of polynomials in k[x].(iii) Every irreducible polynomial of k[x] which has a root in K splits into linear factors in K.296

Def:An extension K=k that satis�es the normality conditions is called a normal extension.297

Thm:Normal extensions remain normal under lifting:(i) If k � E � K and K is normal over k, then K is normal over E.(ii) If K1;K2 are normal over k and are contained in some �eld L, then K1K2 is normal over k and so is K1 \K2.

298

4.3 Separable Extensions

Prop: Let p(x) = IrrPolyk(�) and let � : k ! L be an embedding of k into an algebraically closed �eld L. Thenumber of possible extensions of � to k(�) is � deg p, and is equal to the number of distinct roots of p in ka.299

Def:Given an extension E=F , the separable degree of E over F , written [E : F ]s is the cardinality of the set ofembeddings of E ,! F a over F .300

Thm: \Separable Degrees Multiply" Let k � F � E be a tower. Then [E : k]s = [E : F ]s[F : k]s. Furthermore, if E is�nite over k, then [E : k]s is �nite and [E : k]s � [E : k]. The separable degree is at most equal to the degree.301

Cor: Let E be �nite over k and k � F � E. The equality [E : k]s = [E : k] holds if and only if the correspondingequalities hold for each step of the tower (for E=F and F=k).302

Def:An extension E=k is separable over k if [E : k]s = [E : k]. An element � 2 ka is seperable if k[�] is separable overk. A polynomial f(x) 2 k[x] is separable if it has no multiple roots.303

Thm: Let E be a �nite extension of k. Then E is separable over k if and only if each element of E is separable overk.304

Def:A �eld E is seperable over k if every �nitely generated intermediate extension is separable over k.305

Thm: Let E be an algebraic extension of k, generated by a family of elements f�igi2I . If each �i is separable over k,then E is separable over k.306

Derrick Stolee

4 FIELD THEORY Algebra Study Guide 4.4 Splitting Fields

Def: The compositum of all separable extensions of a �eld k in a given algebraic closure ka is called the separableclosure, denoted ks or ksep.307

Def: If E is an algebraic extension of k and � any embedding of E over k, then �(E) is a conjugate of E in ka. For� 2 ka, if �1; : : : ; �n are distinct embeddings of k[�] into ka, then �(�) is a conjugate of � in ka.308

Def: If a �eld E = k[�], then � is called a primitive element of E over k.309

Thm: \Primitive Element Theorem" Let E be a �nite extension of a �eld k. There exists an element � 2 E such thatE = k[�] if and only if there exists only a �nite number of intermediate �elds F . If E is separable over k, then thereexists such an element �.310

4.4 Splitting Fields

Prop: Let K=F be a splitting �eld of f 2 F [x], let � : F ! F be a �eld isomorphism, and let f 2 F [x] be thecorresponding polynomial. Let K=F be a splitting �eld on f . Then there are [K : F ] distinct isomorphisms K ! Kthat agree with � on F .311

Cor: \Uniqueness of Splitting Fields" If K=F and K=F are splitting �elds of f 2 F [x], then there is an isomorphismK ! K that �xes the elements of F .312

Thm: \Characterization of Galois Extensions" Let K=F be a �eld extension of �nite degree n. The following areequivalent:(i) K=F is Galois (jGal(K=F ) = nj) (ii) jGal(K=F )j � n (iii) KGal(K=F ) = F .(iv) K=F is a splitting �eld of some f 2 F [x]. (v) K=F is a splitting �eld of some irreducible f 2 F [x].(vi) If f is an irreducible polynomial in F [x] and f has at least one root in K, then f splits into linear factors inK[x].313

4.5 Galois Extensions in Characteristic 0

Def: Let K=F be a �eld extension of �nite degree, and let Gal(K=F ) be the Galois group of automorphisms of Kthat �x every element of F . The extension K=F is Galois provided jGal(K=F )j = [K : F ].314

Thm: \Primitive Element Theorem" Given an extension K=F of �nite degree, there exists an � 2 K such that K =F (�).315

Prop: Let K=F be a �eld extension and let f 2 F [x].(i) Suppose � 2 K, f(�) = 0, and � 2 Gal(K=F ). Then f(�(�)) = 0.(ii) Suppose � and � are roots of f in K. Assume f is irreducible over F . Then there is an isomorphism F (�)! F (�)taking � to � and �xing every element of F .316

Thm: Let K be a �eld, and let G be a �nite group of automorphisms of K and let KG = f� 2 K j g(�) = �8g 2 G g.Then K=KG is a Galois extension, and G = Gal(K=KG).317

Thm: \Fundamental Theorem of Galois Theory" Let K=F be a Galois extension. Let L, with F � L � K be anintermediate �eld.(i) For each intermediate �eld L, K=L is Galois, and Gal(K=L) � Gal(K=F ).(ii) For each subgroup H � Gal(K=F ), KH is an intermediate �eld.(iii) The mappings L 7! Gal(K=L) and H 7! KH are reciprocal, order-reversing bijections between intermediate�elds and subgroups of Gal(K=F ).Suppose L is an intermediate �eld with corresponding subgroup Gal(K=L) = H.(iv) [K : L] = jHj and [L : F ] = [Gal(K=F ) : Gal(K=L)].(v) L=F is Galois () Gal(K=L) E Gal(K=F ) () g(L) = L for every g 2 Gal(K=F ).(vi) When the equivalent conditions of (v) are true, the restriction map � : g 7! gjE is a surjection Gal(K=F ) !Gal(L=F ) with kernel Gal(K=L). In particular, Gal(K=F )=Gal(K=L) �= Gal(L=F ).318

4.6 Galois Theory

4.7 Radical Extensions and Solvable Groups

Def:A �eld extension K=F is a radical extension provided there exist positive integers r; n1; : : : ; nr, and elements�1; : : : ; �r 2 K such that �

njj 2 F (�1; : : : ; �j�1) for j = 1; : : : ; r. The numbers nj are called the associated exponents.

These are not necessarily unique.319

Prop: Let E=F be a splitting �eld of xn � 1 over F . Then Gal(E=F ) is abelian.320

Derrick Stolee

NOTES Algebra Study Guide 4.8 Algebraic Closure

Prop: Let K = E(�), where �n 2 E for some integer n, and suppose E contains a primitive nth root of unity �. TheK=E is Galois and Gal(K=E) is abelian.321

Prop: Let E=F be a Galois extension and K=E a radical extension. Then there is an extension L=E sucht that(i) L=F is Galois, (ii) L=E is a radical extension, and (iii) no new exponents are introduced.322

Thm: Let f 2 F [x] be an irreducible polynomial with a root in some radical extension of F . Then the Galois group off is solvable.323

4.8 Algebraic Closure

Def: Let E be an extension �eld of F and let � : F ! L be an embedding. An embedding � : E ! L is said to beover � (or � extends �) if the restriction of � to F is equal to �. If � = idF , then � is an embedding of E over F .324

Lma: Let E be an algebraic extension of k, and let � : E ! E be an embedding of E into itself over k. Then � is anautomorphism.325

Lma: Let E1; E2 be extensions of k, contained in some bigger �eld E and let � be an embedding of E in some �eld L.Then �(E1E2) = �(E1)�(E2).

326

Prop: Let k be a �eld and f a polynomial in k[x] of degree � 1. Then there exists an extension E of k in which f hasa root.327

Cor: Let k be a �eld and let f1; : : : ; fn be polynomials in k[x] of degrees � 1. Then there exists an extension �eld E ofk in which each fi has a root.328

Def:A �eld L is algebraically closed if every polynomial in iL[x] of degree � 1 has a root in L.329

Thm: Let k be a �eld. There exists an algebraically closed �eld with k as a sub�eld.330

Cor: Let k be a �eld. There exists an extension ka which is algebraic over k and algebraically closed.331

Thm: Let k be a �eld, E an algebraic extension of k, and � : k ! L an embedding of k into an algebraically closed �eldL. Then there exists an extension of � to an embedding of E in L. If E is algebraically closed and L is algebraicover �(k), then any such extension of � is an isomorphism of E onto L.332

Cor: Let k be a �eld, and let E;E0 be algebraic extensions of k. If E and E0 are algebraically closed, then there existsan isomorphism � : E ! E0 over k.333

Def: The unique (up to isomorphism) �eld extension of k that is algebraic and is algebraically closed is the algebraicclosure of k, written ka.

4.9 Finite Fields

Thm: For each prime p and integer n � 1 there exists a �nite �eld of order pn denoted Fpn , uniquely determined asa sub�eld of an algebraic closure Fap. It is the splitting �eld of the polynomial xp

n

� x and its elements are the rootsof this polynomial. Every �nite �eld is isomorphic to exactly one �eld Fpn .334

Cor: Let Fq be a �nite �eld. Let n be an integer � 1. In a given algebraic closure Faq , there exists one and only oneextension of Fq of degree n, and this extension is the �eld Fqn .335

Thm: The multiplicative group of a �nite �eld is cyclic.336

Def: Let q be a prime power. Consider the extension Fqn=Fq. The Frobenius map, 'q : Fqn ! Fqn , such that 'q(x) = xq

is an automorphism of Fqn over Fq.337

Thm: The group of automorphisms of Fqn is cyclic of degree n, generated by 'q.338

Thm: Let m;n be integers � 1. Then in any algebraic closure of Fp, the sub�eld Fpn is contained in Fpm if and onlyif n divides m. If that is the case, let q = pn and let m = nd. Then Fqd is normal and separable over Fq, and thegroup of automorphisms of Fpm over Fq is cyclic of order d, generated by 'dp = 'q.

339

Artin: Algebra by Michael Artin.D&F: Abstract Algebra by Dummit & Foote.H: Algebra by Thomas W. Hungerford.Lang: Algebra by Serge Lang.

Notes

262Hungerford. Def 1.1. p. 231. Lang. p.223.263Hungerford. p. 231. Lang p. 224.

264Hungerford. Def. 1.4. p.233; Lang. p.224.265Lang. Prop 1.1. p.224.266Hungerford. p. 231.267Hungerford. Thm. 1.2. p. 231; Lang. Cor. 1.3. p.225.268Lang. Prop. 1.2. p. 224.269Lang. p.225.270Hungerford. p.232; Lang. p. 226.271Lang. Prop. 1.5. p.226.272Hungerford. Thm. 1.3 p. 232.273Hungerford. p. 233; Lang. p.226.274Lang. p.227.

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NOTES Algebra Study Guide NOTES

275Hungerford. p. 233.276Hungerford. Thm 1.5. p.233.277Lang. p.224.278Lang. Prop 1.4. p.225.279Hungerford. Thm. 1.6. p. 234.280Lang. Prop. 1.6. p.227.281Hungerford. Def. 1.7. p.234; Lang. p. 224.282Hungerford. Thm 1.8. p. 235.283Hungerford. Cor 1.9. p.236.284Hungerford. Thm 1.10. p.236.285Hungerford. Thm 1.11. p.237.286Hungerford. Thm.1.12. p.237.287Hungerford. Thm. 1.13. p.237.288Hungerford. Thm. 1.14. p.238.289Lang. p.227.290Lang. Prop 1.7. p.228.291Lang. Thm 4.5. p.241.292Lang. p.235.293Lang. p.236.294Lang. Thm. 3.1. p.236.295Lang. Cor.3.2.p.237.296Lang. Thm 3.3 p.237.297Lang. p.238.298Lang. Thm 3.4 p.238.299Lang. Prop. 2.7. p. 233.300Lang. p.239.301Lang. Thm 4.1 pp.239-240.302Lang Cor 4.2 p.240.303Lang p.240.304Lang Thm 4.3 p.241.305Lang p.241.306Lang Thm 4.4 p.241.307Lang p.243.

308Lang p.243.309Lang p.244.310Lang Thm 4.6. p.243.311R. Wiegand, \Galois Theory Review" Proposition 2.1.312R. Wiegand, \Galois Theory Review" Corollary 2.2.313R. Wiegand, \Galois Theory Review" Theorem 2.3.314R Wiegand. \Galois Theory Review" De�nition 1.1.315R Wiegand. \Galois Theory Review" Theorem 1.2.316R Wiegand. \Galois Theory Review" Proposition 1.3.317R Wiegand. \Galois Theory Review" Theorem 1.5.318R Wiegand. \Galois Theory Review" Theorem 1.5319R. Wiegand, \Galois Theory Review" De�nition 3.2.320R. Wiegand, \Galois Theory Review" Proposition 3.2.321R. Wiegand, \Galois Theory Review" Proposition 3.3.322R. Wiegand, \Galois Theory Review" Proposition 3.4.323R. Wiegand, \Galois Theory Review" Theorem 3.8.324Lang. p.229.325Lang. Lma. 2.1. p.230.326Lang. Lma. 2.2. p.230.327Lang. Prop. 2.3. p.231.328Lang. Cor. 2.4.p.231.329Lang. p.231.330Lang. Thm. 2.5. p.231.331Lang. Cor. 2.6. p.232.332Lang. Thm. 2.8. p.233.333Lang. Cor. 2.9. p.234.334Lang Thm 5.1 p.246.335Lang Cor 5.2 p.246.336Lang Thm 5.3 p.246.337Lang p.246.338Lang Thm 5.4 p.246.339Lang Thm 5.5 p.247.

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5 PROOFS TO KNOW Algebra Study Guide

5 Proofs to Know

Thm: \Hilbert Basis Theorem" Let R be a commutative ring with identity. If R is Noetherian, then so is R[x].340

Proof. Let I be an ideal in R[x]. De�ne a sequence of ideals Ai (i � 0) consisting of 0 and all the elements of Rthat are leading coe�cients of a polynomial in I of degree i. Each Ai is an ideal since R � R[x], and I has closureunder multiplication and addition. Also, Ai � Ai+1 since for each a 2 Ai there is a polynomial fa 2 I of degree iwith leading coe�cient a. Then xfa 2 I as well, has degree i+ 1 and leading coe�cient a, so a 2 Ai+1.

This leads to an ascending chain of ideals A0 � A1 � � � � in R. Since R is Noetherian, this sequence eventually stabilizesat Am for some m � 0, that is An = Am for all n � m. By a theorem, these ideals are �nitely generated. So, letr1; r2; : : : ; rni be generators for Ai, appending new generators for each ascending ideal until Am. If Ai = 0, then setni = 0.

Choose polynomials fj;i 2 I such that fj;i has leading coe�cient ri and is of degree j for all 0 � j � m and 0 < i � nj .I claim that this �nite list of polynomials generates J .

Let X be the ideal generated by the fj;i's. Clearly X � J . Given g 2 J with g 6= 0. Prove by induction on deg g. Casedeg g = 0, then g is a constant, so its leading term is r 2 A0, which has r = s1r1 + � � �+ si0ri0 with s1; : : : ; si0 2 R.Thus, g = s1f0;1 + � � �+ si0f0;i0 2 X. Assume g 2 X for all g with deg g < k.

If k � m, then the leading coe�cient of g is r = s1r1 + � � � + sikrik 2 Ak. Then, g �ikXi=1

sifk;i 2 J has degree strictly

less than k, so must have g �Piki=1 sifk;i 2 X and therefore g 2 X.

If k > m, then the leading coe�cient of g is r = s1r1 + � � � + simrim 2 Ak = Am. Then, g �imXi=1

sifm;ixk�m 2 J has

degree strictly less than k, so is in X. This implies g 2 X.Therefore, J � X, so equality holds and J is �nitely generated. Since every ideal of R[x] is �nitely generated, then R[x]

is Noetherian.

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5 PROOFS TO KNOW Algebra Study Guide

Thm: \Spectral Theorem" Every normal matrix over C is unitarily similar to a diagonal matrix. [Recall a square matrixP is unitary given PP � = I. A square matrix P is normal if PP � = P �P . Two square matrices are unitarily similarif there is a unitary matrix P such that P �AP = B.]

Proof. Proceed by induction on n. When n = 1, a normal matrix N has a single entry, so is diagonal. Assume everynormal matrix over C of size n� n is unitarily similar to a diagonal matrix for n < k for some k > 1.

Let A be a k � k normal matrix. Let v be an eigenvector for an eigenvalue �. Without loss of generality, assume v haslength 1 by resizing. v can be extended to an orthonormal basis fv; v2; : : : ; vkg. Let Q =

�v v2 � � � vk

�2 Uk, a

unitary matrix. Then, de�ne

A0 = Q�AQ =

26664�1 a2 � � � ak0... A1

0

37775 ;

where A1 is a (k � 1)� (k � 1) matrix acting as a minor of A0. Note that

(A0)�A0 = Q�A�QQ�AQ

= Q�A�AQ

= Q�AA�Q (normality of A)

= Q�AQQ�A�Q

= A0(A0)�:

Therefore, A0 is normal. Now consider

"�1�1 � � �... A�1A1

#= (A0)�(A0)

= (A0)(A0)� =

"�1�1 + a1a1 + � � �+ akak � � �

... A1A�1

#

Therefore, �1�1 + a1a1 + � � �+ akak = �1�1. Since aiai 2 R�0, then each ai = 0. So,

A0 =

26664�1 0 � � � 00... A1

0

37775 :

Now, A1 is a normal matrix since"�1�1 � � �... A�1A1

#=

��1

A�1

� ��1

A1

�= (A0)�A0 = A0(A0)� =

��1

A�1

� ��1

A1

�=

"�1�1 � � �... A1A

�1

#:

Thus, A�1A1 = A1A�1 so A1 is normal. By induction, there is a unitary matrix P 2 Uk�1 such that P �A1P = A2

a diagonal matrix. So, Let P 0 =

�1

P

�2 Uk. Then P 0�A0P 0 = P 0�Q�AQP 0 = (QP 0)�A(QP 0) =

��1

A2

�, a

diagonal matrix. Therefore, A is unitarily similar to a diagonal matrix.

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5 PROOFS TO KNOW Algebra Study Guide

Prop: There are no simple groups of order pq where p and q are distinct primes.

Proof. Let G be a group of order pq where p and q are distinct primes. Assume p < q. Consider any Sylow q-subgroupQ. Since [G : Q] = p, the smallest prime divisor of G, then Q E G.

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